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100 Y on a FB F?
100 Yards on a FootBall Field
How do you solve this problem using derivatives find two real numbers x and y so that x-y equals 100 and their product is minimized?
At first glance, this would appear to be a multivariable problem. But in actuality, with the two equations we have to work with: f(x,y) = xy x - y = 100 We can rearrange the second one to find what y is equal to in terms of x: y = 100 - x (We can also solve x in terms of y, but solving y in terms of x is conventional). then we can plug this in for y into f(x,y) to make it simply a function of x: f(x) = x(100 - x) = 100x - x2 We then take the derivative of this, setting f'(x) equal to 0: f'(x) = 0 = 100 - 2x 2x = 100 x = 100/2 = 50 This is the value of x that minimizes the product of x and y when x - y = 100. We then solve for y by plugging in the x-value into this equation: 50 - y = 100 -y = 100 - 50 y = -50 By plugging in values for which x > 50 and x < 50 into the equation f(x) = 100x - x2, we find that f(x) is indeed greater on both the intervals where x < 50 and where x > 50, proving that x = 50 is indeed an absolute minimum.
When was Pierre De fermat's last theorem created?
PIERRE DE FERMAT's last Theorem. (x,y,z,n) belong ( N+ )^4.. n>2. (a) belong Z F is function of ( a.) F(a)=[a(a+1)/2]^2 F(0)=0 and F(-1)=0. Consider two equations F(z)=F(x)+F(y) F(z-1)=F(x-1)+F(y-1) We have a string inference F(z)=F(x)+F(y) equivalent F(z-1)=F(x-1)+F(y-1) F(z)=F(x)+F(y) infer F(z-1)=F(x-1)+F(y-1) F(z-x-1)=F(x-x-1)+F(y-x-1) infer F(z-x-2)=F(x-x-2)+F(y-x-2) we see F(z-x-1)=F(x-x-1)+F(y-x-1 ) F(z-x-1)=F(-1)+F(y-x-1 ) F(z-x-1)=0+F(y-x-1 ) give z=y and F(z-x-2)=F(x-x-2)+F(y-x-2) F(z-x-2)=F(-2)+F(y-x-2) F(z-x-2)=1+F(y-x-2) give z=/=y. So F(z-x-1)=F(x-x-1)+F(y-x-1) don't infer F(z-x-2)=F(x-x-2)+F(y-x-2) So F(z)=F(x)+F(y) don't infer F(z-1)=F(x-1)+F(y-1) So F(z)=F(x)+F(y) is not equivalent F(z-1)=F(x-1)+F(y-1) So have two cases. [F(x)+F(y)] = F(z) and F(x-1)+F(y-1)]=/=F(z-1) or vice versa So [F(x)+F(y)]-[F(x-1)+F(y-1)]=/=F(z)-F(z-1). Or F(x)-F(x-1)+F(y)-F(y-1)=/=F(z)-F(z-1). We have F(x)-F(x-1) =[x(x+1)/2]^2 - [(x-1)x/2]^2. =(x^4+2x^3+x^2/4) - (x^4-2x^3+x^2/4). =x^3. F(y)-F(y-1) =y^3. F(z)-F(z-1) =z^3. So x^3+y^3=/=z^3. n>2. .Similar. We have a string inference G(z)*F(z)=G(x)*F(x)+G(y)*F(y) equivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z)=G(x)*F(x)+G(y)*F(y) infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) we see G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=G(x)*F(-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=0+G(y)*F(y-x-1 ) give z=y. and G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)*F(-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)+G(y)*F(y-x-2) x>0 infer G(x)>0. give z=/=y. So G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) don't infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) So G(z)*F(z)=G(x)*F(x)+G(y)*F(y) don't infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) So G(z)*F(z)=G(x)*F(x)+G(y)*F(y) is not equiivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) So have two cases [G(x)*F(x)+G(y)*F(y)]=G(z)*F(z) and [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z-1)*F(z-1) or vice versa. So [G(x)*F(x)+G(y)*F(y)] - [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z)*[F(z)-F(z-1)]. Or G(x)*[F(x) - F(x-1)] + G(y)*[F(y)-F(y-1)]=/=G(z)*[F(z)-F(z-1).] We have x^n=G(x)*[F(x)-F(x-1) ] y^n=G(y)*[F(y)-F(y-1) ] z^n=G(z)*[F(z)-F(z-1) ] So x^n+y^n=/=z^n Happy&Peace. Trần Tấn Cường.
What is the derivative of yXy'?
assuming that you are referring to a function: f(y) = y×y or better put: f(y) = y2 Then it's derivative would be: f'(y) = 2y
What notation represents a function as f x instead of y?
'Y' is a function 'f' of 'x': Y = f(x) . 'Z' is a function 'g' of 'y': Z = g [ f(x) ] .
What does 100 y in a f f mean?
100 yards in american football fields.
100 Y on a FB F?
100 Yards on a FootBall Field
How do you solve this problem using derivatives find two real numbers x and y so that x-y equals 100 and their product is minimized?
At first glance, this would appear to be a multivariable problem. But in actuality, with the two equations we have to work with: f(x,y) = xy x - y = 100 We can rearrange the second one to find what y is equal to in terms of x: y = 100 - x (We can also solve x in terms of y, but solving y in terms of x is conventional). then we can plug this in for y into f(x,y) to make it simply a function of x: f(x) = x(100 - x) = 100x - x2 We then take the derivative of this, setting f'(x) equal to 0: f'(x) = 0 = 100 - 2x 2x = 100 x = 100/2 = 50 This is the value of x that minimizes the product of x and y when x - y = 100. We then solve for y by plugging in the x-value into this equation: 50 - y = 100 -y = 100 - 50 y = -50 By plugging in values for which x > 50 and x < 50 into the equation f(x) = 100x - x2, we find that f(x) is indeed greater on both the intervals where x < 50 and where x > 50, proving that x = 50 is indeed an absolute minimum.
When was Pierre De fermat's last theorem created?
PIERRE DE FERMAT's last Theorem. (x,y,z,n) belong ( N+ )^4.. n>2. (a) belong Z F is function of ( a.) F(a)=[a(a+1)/2]^2 F(0)=0 and F(-1)=0. Consider two equations F(z)=F(x)+F(y) F(z-1)=F(x-1)+F(y-1) We have a string inference F(z)=F(x)+F(y) equivalent F(z-1)=F(x-1)+F(y-1) F(z)=F(x)+F(y) infer F(z-1)=F(x-1)+F(y-1) F(z-x-1)=F(x-x-1)+F(y-x-1) infer F(z-x-2)=F(x-x-2)+F(y-x-2) we see F(z-x-1)=F(x-x-1)+F(y-x-1 ) F(z-x-1)=F(-1)+F(y-x-1 ) F(z-x-1)=0+F(y-x-1 ) give z=y and F(z-x-2)=F(x-x-2)+F(y-x-2) F(z-x-2)=F(-2)+F(y-x-2) F(z-x-2)=1+F(y-x-2) give z=/=y. So F(z-x-1)=F(x-x-1)+F(y-x-1) don't infer F(z-x-2)=F(x-x-2)+F(y-x-2) So F(z)=F(x)+F(y) don't infer F(z-1)=F(x-1)+F(y-1) So F(z)=F(x)+F(y) is not equivalent F(z-1)=F(x-1)+F(y-1) So have two cases. [F(x)+F(y)] = F(z) and F(x-1)+F(y-1)]=/=F(z-1) or vice versa So [F(x)+F(y)]-[F(x-1)+F(y-1)]=/=F(z)-F(z-1). Or F(x)-F(x-1)+F(y)-F(y-1)=/=F(z)-F(z-1). We have F(x)-F(x-1) =[x(x+1)/2]^2 - [(x-1)x/2]^2. =(x^4+2x^3+x^2/4) - (x^4-2x^3+x^2/4). =x^3. F(y)-F(y-1) =y^3. F(z)-F(z-1) =z^3. So x^3+y^3=/=z^3. n>2. .Similar. We have a string inference G(z)*F(z)=G(x)*F(x)+G(y)*F(y) equivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z)=G(x)*F(x)+G(y)*F(y) infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) we see G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=G(x)*F(-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=0+G(y)*F(y-x-1 ) give z=y. and G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)*F(-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)+G(y)*F(y-x-2) x>0 infer G(x)>0. give z=/=y. So G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) don't infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) So G(z)*F(z)=G(x)*F(x)+G(y)*F(y) don't infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) So G(z)*F(z)=G(x)*F(x)+G(y)*F(y) is not equiivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) So have two cases [G(x)*F(x)+G(y)*F(y)]=G(z)*F(z) and [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z-1)*F(z-1) or vice versa. So [G(x)*F(x)+G(y)*F(y)] - [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z)*[F(z)-F(z-1)]. Or G(x)*[F(x) - F(x-1)] + G(y)*[F(y)-F(y-1)]=/=G(z)*[F(z)-F(z-1).] We have x^n=G(x)*[F(x)-F(x-1) ] y^n=G(y)*[F(y)-F(y-1) ] z^n=G(z)*[F(z)-F(z-1) ] So x^n+y^n=/=z^n Happy&Peace. Trần Tấn Cường.
Who can solve FLT?
Последнее Пьер де Ферма теоремы. (x,y,z,n) принадлежать( N+ )^4. n>2. (a) принадлежать Z F является функцией( a.) F(a)=[a(a+1)/2]^2 F(0)=0 и F(-1)=0. Рассмотрим два уравнения F(z)=F(x)+F(y) F(z-1)=F(x-1)+F(y-1) непрерывный дедуктивного рассуждения F(z)=F(x)+F(y) эквивалент F(z-1)=F(x-1)+F(y-1) F(z)=F(x)+F(y) выводить F(z-1)=F(x-1)+F(y-1) F(z-x-1)=F(x-x-1)+F(y-x-1) выводить F(z-x-2)=F(x-x-2)+F(y-x-2) мы видим, F(z-x-1)=F(x-x-1)+F(y-x-1 ) F(z-x-1)=F(-1)+F(y-x-1 ) F(z-x-1)=0+F(y-x-1 ) давать z=y и F(z-x-2)=F(x-x-2)+F(y-x-2) F(z-x-2)=F(-2)+F(y-x-2) F(z-x-2)=1+F(y-x-2) давать z=/=y. так F(z-x-1)=F(x-x-1)+F(y-x-1) не выводить F(z-x-2)=F(x-x-2)+F(y-x-2) так F(z)=F(x)+F(y) не выводить F(z-1)=F(x-1)+F(y-1) так F(z)=F(x)+F(y) не эквивалентен F(z-1)=F(x-1)+F(y-1) Таким образом, возможны два случая. [F(x)+F(y)] = F(z) и F(x-1)+F(y-1)]=/=F(z-1) или наоборот так [F(x)+F(y)]-[F(x-1)+F(y-1)]=/=F(z)-F(z-1). или F(x)-F(x-1)+F(y)-F(y-1)=/=F(z)-F(z-1). у нас есть F(x)-F(x-1) =[x(x+1)/2]^2 - [(x-1)x/2]^2. =(x^4+2x^3+x^2/4) - (x^4-2x^3+x^2/4). =x^3. F(y)-F(y-1) =y^3. F(z)-F(z-1) =z^3. так x^3+y^3=/=z^3. n>2. аналогичный непрерывный дедуктивного рассуждения G(z)*F(z)=G(x)*F(x)+G(y)*F(y) эквивалент G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z)=G(x)*F(x)+G(y)*F(y) выводить G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) выводить G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) мы видим, G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=G(x)*F(-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=0+G(y)*F(y-x-1 ) давать z=y. и G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)*F(-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)+G(y)*F(y-x-2) x>0 выводить G(x)>0. давать z=/=y. так G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y)не выводить G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) так G(z)*F(z)=G(x)*F(x)+G(y)*F(y) не выводить G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) так G(z)*F(z)=G(x)*F(x)+G(y)*F(y) не эквивалентен G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) Таким образом, возможны два случая. [G(x)*F(x)+G(y)*F(y)]=G(z)*F(z) и [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z-1)*F(z-1) или наоборот. так [G(x)*F(x)+G(y)*F(y)] - [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z)*[F(z)-F(z-1)]. или G(x)*[F(x) - F(x-1)] + G(y)*[F(y)-F(y-1)]=/=G(z)*[F(z)-F(z-1).] у нас есть x^n=G(x)*[F(x)-F(x-1) ] y^n=G(y)*[F(y)-F(y-1) ] z^n=G(z)*[F(z)-F(z-1) ] так x^n+y^n=/=z^n Счастливые и мира. Trần Tấn Cường.
What is shortest solve about fermat?
To: trantancuong21@yahoo.com PIERRE DE FERMAT's last Theorem. (x,y,z,n) belong ( N+ )^4.. n>2. (a) belong Z F is function of ( a.) F(a)=[a(a+1)/2]^2 F(0)=0 and F(-1)=0. Consider two equations F(z)=F(x)+F(y) F(z-1)=F(x-1)+F(y-1) We have a string inference F(z)=F(x)+F(y) equivalent F(z-1)=F(x-1)+F(y-1) F(z)=F(x)+F(y) infer F(z-1)=F(x-1)+F(y-1) F(z-x-1)=F(x-x-1)+F(y-x-1) infer F(z-x-2)=F(x-x-2)+F(y-x-2) we see F(z-x-1)=F(x-x-1)+F(y-x-1 ) F(z-x-1)=F(-1)+F(y-x-1 ) F(z-x-1)=0+F(y-x-1 ) give z=y and F(z-x-2)=F(x-x-2)+F(y-x-2) F(z-x-2)=F(-2)+F(y-x-2) F(z-x-2)=1+F(y-x-2) give z=/=y. So F(z-x-1)=F(x-x-1)+F(y-x-1) don't infer F(z-x-2)=F(x-x-2)+F(y-x-2) So F(z)=F(x)+F(y) don't infer F(z-1)=F(x-1)+F(y-1) So F(z)=F(x)+F(y) is not equivalent F(z-1)=F(x-1)+F(y-1) So have two cases. [F(x)+F(y)] = F(z) and F(x-1)+F(y-1)]=/=F(z-1) or vice versa So [F(x)+F(y)]-[F(x-1)+F(y-1)]=/=F(z)-F(z-1). Or F(x)-F(x-1)+F(y)-F(y-1)=/=F(z)-F(z-1). We have F(x)-F(x-1) =[x(x+1)/2]^2 - [(x-1)x/2]^2. =(x^4+2x^3+x^2/4) - (x^4-2x^3+x^2/4). =x^3. F(y)-F(y-1) =y^3. F(z)-F(z-1) =z^3. So x^3+y^3=/=z^3. n>2. .Similar. We have a string inference G(z)*F(z)=G(x)*F(x)+G(y)*F(y) equivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z)=G(x)*F(x)+G(y)*F(y) infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) we see G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=G(x)*F(-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=0+G(y)*F(y-x-1 ) give z=y. and G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)*F(-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)+G(y)*F(y-x-2) x>0 infer G(x)>0. give z=/=y. So G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) don't infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) So G(z)*F(z)=G(x)*F(x)+G(y)*F(y) don't infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) So G(z)*F(z)=G(x)*F(x)+G(y)*F(y) is not equiivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) So have two cases [G(x)*F(x)+G(y)*F(y)]=G(z)*F(z) and [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z-1)*F(z-1) or vice versa. So [G(x)*F(x)+G(y)*F(y)] - [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z)*[F(z)-F(z-1)]. Or G(x)*[F(x) - F(x-1)] + G(y)*[F(y)-F(y-1)]=/=G(z)*[F(z)-F(z-1).] We have x^n=G(x)*[F(x)-F(x-1) ] y^n=G(y)*[F(y)-F(y-1) ] z^n=G(z)*[F(z)-F(z-1) ] So x^n+y^n=/=z^n Happy&Peace. Trần Tấn Cường.
What is the derivative of yXy'?
assuming that you are referring to a function: f(y) = y×y or better put: f(y) = y2 Then it's derivative would be: f'(y) = 2y
What is new solve about Fermat?
To: trantancuong21@yahoo.com Последнее Пьер де Ферма теоремы. . (x,y,z,n) принадлежать( N+ )^4.. n>2. (a) принадлежать Z F является функцией( a.) F(a)=[a(a+1)/2]^2 F(0)=0 и F(-1)=0. Рассмотрим два уравнения F(z)=F(x)+F(y) F(z-1)=F(x-1)+F(y-1) непрерывный дедуктивного рассуждения F(z)=F(x)+F(y) эквивалент F(z-1)=F(x-1)+F(y-1) F(z)=F(x)+F(y) выводить F(z-1)=F(x-1)+F(y-1) F(z-x-1)=F(x-x-1)+F(y-x-1) выводить F(z-x-2)=F(x-x-2)+F(y-x-2) мы видим, F(z-x-1)=F(x-x-1)+F(y-x-1 ) F(z-x-1)=F(-1)+F(y-x-1 ) F(z-x-1)=0+F(y-x-1 ) давать z=y и F(z-x-2)=F(x-x-2)+F(y-x-2) F(z-x-2)=F(-2)+F(y-x-2) F(z-x-2)=1+F(y-x-2) давать z=/=y. так F(z-x-1)=F(x-x-1)+F(y-x-1) не выводить F(z-x-2)=F(x-x-2)+F(y-x-2) так F(z)=F(x)+F(y) не выводить F(z-1)=F(x-1)+F(y-1) так F(z)=F(x)+F(y) не эквивалентен F(z-1)=F(x-1)+F(y-1) Таким образом, возможны два случая. [F(x)+F(y)] = F(z) и F(x-1)+F(y-1)]=/=F(z-1) или наоборот так [F(x)+F(y)]-[F(x-1)+F(y-1)]=/=F(z)-F(z-1). или F(x)-F(x-1)+F(y)-F(y-1)=/=F(z)-F(z-1). у нас есть F(x)-F(x-1) =[x(x+1)/2]^2 - [(x-1)x/2]^2. =(x^4+2x^3+x^2/4) - (x^4-2x^3+x^2/4). =x^3. F(y)-F(y-1) =y^3. F(z)-F(z-1) =z^3. так x^3+y^3=/=z^3. n>2. аналогичный непрерывный дедуктивного рассуждения G(z)*F(z)=G(x)*F(x)+G(y)*F(y) эквивалент G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z)=G(x)*F(x)+G(y)*F(y) выводить G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) выводить G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) мы видим, G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=G(x)*F(-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=0+G(y)*F(y-x-1 ) давать z=y. и G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)*F(-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)+G(y)*F(y-x-2) x>0 выводить G(x)>0. давать z=/=y. так G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y)не выводить G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) так G(z)*F(z)=G(x)*F(x)+G(y)*F(y) не выводить G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) так G(z)*F(z)=G(x)*F(x)+G(y)*F(y) не эквивалентен G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) Таким образом, возможны два случая. [G(x)*F(x)+G(y)*F(y)]=G(z)*F(z) и [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z-1)*F(z-1) или наоборот. так [G(x)*F(x)+G(y)*F(y)] - [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z)*[F(z)-F(z-1)]. или G(x)*[F(x) - F(x-1)] + G(y)*[F(y)-F(y-1)]=/=G(z)*[F(z)-F(z-1).] у нас есть x^n=G(x)*[F(x)-F(x-1) ] y^n=G(y)*[F(y)-F(y-1) ] z^n=G(z)*[F(z)-F(z-1) ] так x^n+y^n=/=z^n Счастливые и мира. Trần Tấn Cường.
Who do read my FLT 's solve?
Địng lý cuối của PIERRE DE FERMAT. (x,y,z,n) thuộc tập hợp ( N+ )^4.. n>2. (a) thuộc tập hợp Z F là hàm số của ( a.) F(a)=[a(a+1)/2]^2 F(0)=0 и F(-1)=0. Xét hai phương trình F(z)=F(x)+F(y) F(z-1)=F(x-1)+F(y-1) Ta có một dãy suy luận F(z)=F(x)+F(y) tương đương F(z-1)=F(x-1)+F(y-1) F(z)=F(x)+F(y) suy ra F(z-1)=F(x-1)+F(y-1) F(z-x-1)=F(x-x-1)+F(y-x-1) suy ra F(z-x-2)=F(x-x-2)+F(y-x-2) ta có F(z-x-1)=F(x-x-1)+F(y-x-1 ) F(z-x-1)=F(-1)+F(y-x-1 ) F(z-x-1)=0+F(y-x-1 ) cho z=y và F(z-x-2)=F(x-x-2)+F(y-x-2) F(z-x-2)=F(-2)+F(y-x-2) F(z-x-2)=1+F(y-x-2) cho z=/=y. do đó F(z-x-1)=F(x-x-1)+F(y-x-1) không suy raF(z-x-2)=F(x-x-2)+F(y-x-2) do đó F(z)=F(x)+F(y) không suy ra F(z-1)=F(x-1)+F(y-1) do đó F(z)=F(x)+F(y) không tương đương F(z-1)=F(x-1)+F(y-1) điều có thể xảy ra [F(x)+F(y)] = F(z) и F(x-1)+F(y-1)]=/=F(z-1) hay ngược lại [F(x)+F(y)]-[F(x-1)+F(y-1)]=/=F(z)-F(z-1). hoặc F(x)-F(x-1)+F(y)-F(y-1)=/=F(z)-F(z-1). ta có F(x)-F(x-1) =[x(x+1)/2]^2 - [(x-1)x/2]^2. =(x^4+2x^3+x^2/4) - (x^4-2x^3+x^2/4). =x^3. F(y)-F(y-1) =y^3. F(z)-F(z-1) =z^3. do đó x^3+y^3=/=z^3. n>2.tương tự Ta có một dãy suy luận F(z)=G(x)*F(x)+G(y)*F(y) tương đương G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z)=G(x)*F(x)+G(y)*F(y) suy ra G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) suy ra G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) ta có G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=G(x)*F(-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=0+G(y)*F(y-x-1 ) cho z=y. và G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)*F(-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)+G(y)*F(y-x-2) x>0 do đó G(x)>0. cho z=/=y. do đó G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y)не выводить G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) do đó G(z)*F(z)=G(x)*F(x)+G(y)*F(y) не выводить G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) do đó G(z)*F(z)=G(x)*F(x)+G(y)*F(y) не эквивалентен G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) điều có thể xảy ra [G(x)*F(x)+G(y)*F(y)]=G(z)*F(z) и [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z-1)*F(z-1) hay ngược lại do đó [G(x)*F(x)+G(y)*F(y)] - [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z)*[F(z)-F(z-1)]. hay G(x)*[F(x) - F(x-1)] + G(y)*[F(y)-F(y-1)]=/=G(z)*[F(z)-F(z-1).] ta có x^n=G(x)*[F(x)-F(x-1) ] y^n=G(y)*[F(y)-F(y-1) ] z^n=G(z)*[F(z)-F(z-1) ] do đó x^n+y^n=/=z^n Hòa Bình. Trần Tấn Cường.
What is grace solve about Fermat?
Pierre de Fermat's laatste stelling (x, y, z, n) element van de (N +) ^ 4 n> 2 (a) element van de Z F is de functie (s). F (a) = [a (a +1) / 2] ^ 2 F (0) = 0 en F (-1) = 0 Beschouw twee vergelijkingen. F (z) = F (x) + F (y) F (z-1) = F (x-1) + F (y-1) We hebben een keten van gevolgtrekking F (z) = F (x) + F (y) gelijkwaardig F (z-1) = F (x-1) + F (y-1) F (z) = F (x) + F (y) conclusie F (z-1) = F (x-1) + F (y-1) F (z-x-1) = F (x-x-1) + F (y-x-1) conclusie F (z-x-2) = F (x-x-2) + F (y-x-2) zien we F (z-x-1) = F (x-x-1) + F (y-x-1) F (z-x-1) = F (-1) + F (y-x-1) F (z-x-1) = 0 + F (y-x-1) conclusie z = y en F (z-x-2) = F (x-x-2) + F (y-x-2) F (z-x-2) = F (-2) + F (y-x-2) F (z-x-2) = 1 + F (y-x-2) conclusie z = / = y. conclusie F (z-x-1) = F (x-x-1) + F (y-x-1) geen conclusie (z-x-2) = F (x-x 2) + F (y-x-2) conclusie F (z) = F (x) + F (y) geen conclusie F (z-1) = F (x-1) + F (y-1) conclusie F (z) = F (x) + F (y) zijn niet equivalent van F (z-1) = F (x-1) + F (y-1) Daarom is de twee gevallen. [F (x) + F (y)] = F (z) en F (x-1) + F (y-1)] = / = F (Z-1) of vice versa conclusie [F (x) + F (y)] - [F (x-1) + F (y-1)] = / = F (z) - F (z-1). of F (x) - F (x-1) + F (y)-F (y-1) = / = F (z) - F (z-1). zien we F (x) - F (x-1) = [x (x 1) / 2] ^ 2 - [(x-1) x / 2] ^ 2 = (X ^ 4 +2 x ^ 3 + x ^ 2/4) - (x ^ 4-2x 3 + x ^ ^ 2/4). = X ^ 3 F (y)-F (y-1) = y ^ 3 F (z)-F (z-1) = z ^ 3 conclusie x 3 + y ^ 3 = / = z ^ 3 n> 2. lossen soortgelijke We hebben een keten van gevolgtrekking G (z) * F (z) = G (x) * F (x) + G (y) * F (y) gelijkwaardig G (z) * F (z-1) = G (x) * F ( x -1) + G (y) * F (y-1) G (z) * F (z) = G (x) * F (x) + G (y) * F (y) conclusie G (z) * F (z-1) = G (x) * F (x -1) + G (y) * F (y-1) G (z) * F (z-x-1) = G (x) * F (x-x-1) + G (y-x-1) * F (y) conclusie G (z) * F (z-x-2) = G ( x) * F (x-x 2) + G (y) * F (y-x 2) zien we G (z) * F (z-x-1) = G (x) * F (x-x-1) + G (y) * F (y-x-1) G (z) * F (z-x-1) = G (x) * F (-1) + G (y) * F (y-x-1) G (z) * F (z-x-1) = 0 + G (y) * F (y-x-1) conclusie z = y. en G (z) * F (z-x-2) = G (x) * F (x-x-2) + G (y) * F (y-x-2) G (z) * F (z-x-2) = G (x) * F (-2) + G (y) * F (y-x-2) G (z) * F (z-x-2) = G (x) + G (y) * F (x-y-2) x> 0 conclusie G (x)> 0 conclusie z = / = y. conclusie G (z) * F (z-x-1) = G (x) * F (x-x-1) + G (y-x-1) * F (y) geen conclusie G (z) * F (z-x-2) = G (x) * F (x-x-2) + G (y) * F (y-x-2) conclusie G (z) * F (z) = G (x) * F (x) + G (y) * F (y) geen conclusie G (z) * F (z-1) = G (x) * F ( x-1) + G (y) * F (y-1) conclusie G (z) * F (z) = G (x) * F (x) + G (y) * F (y) zijn niet equivalent van G (z) * F (z-1) = G (x) * F (x-1) + G (y) * F (y-1) Daarom is de twee gevallen. [G (x) * F (x) + G (y) * F (y)] = G (z) * F (z) en [G (x) * F (x-1) + G (y) * F (y-1)] = / = G (z-1) * F (z-1) of vice versa conclusie [G (x) * F (x) + G (y) * F (y)] - [G (x) * F (x-1) + G (y) * F (y-1)] = / = G (z) * [F (z) -F(Z-1)]. of G (x) * [F (x) - F (x-1)] + G (y) * [F (y) - F (y-1)] = / = G (z) * [F (z) - F(z-1)] zien we x ^ n = G (x) * [F (x) - F (x-1)] y ^ n = G (y) * [F (y) - F (y-1)] z ^ n = G (z) * [F (z) - F (z-1)] conclusie x ^ n + y ^ n = / = z ^ n gelukkig en vrede Tran tan Cuong .
Who do understand Fermat?
ultimo teorema di Pierre De Fermat (x,y,z, n) elemento della (N +)^ 4 n> 2 (a) elemento della Z F è la funzione (a). F (a) = [a(a +1) / 2] ^2 F (0) = 0 e F (-1) = 0 Si considerino due equazioni. F (z) = F (x) + F (y) F (z-1) = F (x-1) + F (y-1) Abbiamo una catena di inferenza F (z) = F (x) + F (y) equivalente F (z-1) = F (x-1) + F (y-1) F (z) = F (x) + F (y) conclusione F (z-1) = F (x-1) + F (y-1) F (z-x-1) = F (x-x-1) + F (y-x-1) conclusione F (z-x-2) = F (x-x-2) + F (y-x-2) vediamo F (z-x-1) = F (x-x-1) + F (y-x-1) F (z-x-1) = F (-1) + F (y-x-1) F (z-x-1) = 0 + F (y-x-1) conclusione z = y e F (z-x-2) = F (x-x-2) + F (y-x-2) F (z-x-2) = F (-2) + F (y-x-2) F (z-x-2) = 1 + F (y-x-2) conclusione z = / = y. conclusione F (z-x-1) = F (x-x-1) + F (y-x-1) alcuna conclusione (z-x-2) = F (x-x 2) + F (y-x-2) conclusione F (z) = F (x) + F (y) alcuna conclusione F (z-1) = F (x-1) + F (y-1) conclusione F (z) = F (x) + F (y) non sono equivalenti di F (z-1) = F (x-1) + F (y-1) Pertanto, i due casi. [F (x) + F (y)] = F (z) e F (x-1) + F (y-1)] = / = F (Z-1) o viceversa conclusione [F (x) + F (y)] - [F (x-1) + F (y-1)] = / = F (z)- F (z-1). Or. F (x)- F (x-1) + F (y) -F (y-1) = / = F (z)- F (z-1). vediamo F (x)- F (x-1) = [x (x 1) / 2] ^ 2 - [(x-1) x / 2] ^2 = (X ^ 4 +2 x ^ 3 + x ^ 2/4) - (x ^ 4-2x ^ 3 + x ^ 2/4). = X ^ 3 F (y) -F (y-1) = y ^ 3 F (z) -F (z-1) = z ^ 3 conclusione x 3 + y ^ 3 =/= z^ 3 n> 2. risolvere simili Abbiamo una catena di inferenza G (z) * F (z) = G (x) * F (x) + G (y) * F (y) equivalenti di G (z) * F (z-1) = G (x) * F ( x -1) + G (y) * F (y-1) G (z) * F (z) = G (x) * F (x) + G (y) * F (y) conclusione G (z) * F (z-1) = G (x) * F (x -1) + G (y) * F (y-1) G (z) * F (z-x-1) = G (x) * F (x-x-1) + G (y-x-1) * F (y) conclusione G (z) * F (z-x-2) = G ( x) * F (x-x 2) + G (y) * F (y-x 2) vediamo G (z) * F (z-x-1) = G (x) * F (x-x-1) + G (y) * F (y-x-1) G (z) * F (z-x-1) = G (x) * F (-1) + G (y) * F (y-x-1) G (z) * F (z-x-1) = 0 + G (y) * F (y-x-1) conclusione z = y. e G (z) * F (z-x-2) = G (x) * F (x-x-2) + G (y) * F (y-x-2) G (z) * F (z-x-2) = G (x) * F (-2) + G (y) * F (y-x-2) G (z) * F (z-x-2) = G (x) + G (y) * F (x-y-2) x> 0 conclusioni G (x)> 0 conclusione z = / = y. conclusione G (z) * F (z-x-1) = G (x) * F (x-x-1) + G (y-x-1) * F (y) alcuna conclusione G (z) * F (z-x-2) = G (x) * F (x-x-2) + G (y) * F (y-x-2) conclusione G (z) * F (z) = G (x) * F (x) + G (y) * F (y) alcuna conclusione G (z) * F (z-1) = G (x) * F ( x-1) + G (y) * F (y-1) conclusione G (z) * F (z) = G (x) * F (x) + G (y) * F (y) non sono equivalenti di G (z) * F (z-1) = G (x) * F ( x-1) + G (y) * F (y-1) Pertanto, i due casi. [G (x) * F (x) + G (y) * F (y)] = G (z) * F (z) e [G (x) * F (x-1) + G (y) * F (y-1)] = / = G (z-1) * F (z-1) o viceversa conclusione [G (x) * F (x) + G (y) * F (y)] - [G (x) * F (x-1) + G (y) * F (y-1)] = / = G (z) * [F (z)- F(Z-1)]. o G (x) * [F (x) - F (x-1)] + G (y) * [F (y)- F (y-1)] = / = G (z) * [F (z)- F(z-1).] vediamo x^ n = G (x) * [F (x)- F (x-1)] y ^ n = G (y) * [F (y)- F (y-1)] z ^ n = G (z) * [F (z)- F (z-1)] conclusione x ^ n + y ^ n = / = z ^ n Felicità e la pace Cuong Tran
Who can solve FLT short?
Le dernier théorème de Pierre de Fermat . (x, y, z, n) l'ensemble de ( N+ )^4. n> 2. ( a ) l'ensemble de Z F est la fonction de (a). F (a) = [a (a +1) / 2] ^ 2 F (0) = 0 et F (-1) = 0. Considérons deux équations. F (z) = F (x) + F (y) F (z-1) = F (x-1) + F (y-1) Nous avons une inférence chaîne F (z) = F (x) + F (y) équivalent F (z-1) = F (x-1) + F (y-1) F (z) = F (x) + F (y) en déduire F (z-1) = F (x-1) + F (y-1) F (z-x-1) = F (x-x-1) + F (y-x-1) en déduire F (z-x-2) = F (x-x-2) + F (y-x-2) nous voyons F (z-x-1) = F (x-x-1) + F (y-x-1) F (z-x-1) = F (-1) + F (y-x-1) F (z-x-1) = 0 + F (y-x-1) donner z = y et F (z-x-2) = F (x-x-2) + F (y-x-2) F (z-x-2) = F (-2) + F (y-x-2) F (z-x-2) = 1 + F (y-x-2) donner z = / = y. de sorte F (z-x-1) = F (x-x-1) + F (y-x-1) ne pas en déduire F (z-x-2) = F (x-x-2) + F (y-x-2) de sorte F (z) = F (x) + F (y) ne pas en déduire F (z-1) = F (x-1) + F (y-1) de sorte F (z) = F (x) + F (y) n'est pas équivalente F (z-1) = F (x-1) + F (y-1) Donc avoir deux cas. [F (x) + F (y)] = F (z) et F (x-1) + F (y-1)] = / = F (z-1) ou vice versa de sorte [F (x) + F (y)] - [F (x-1) + F (y-1)] = / = F (z)-F (z-1). Ou F (x)-F (x-1) + F (y)-F (y-1) = / = F (z)-F (z-1). nous voyons F(x)-F(x-1) =[x(x+1)/2]^2 - [(x-1)x/2]^2. =(x^4+2x^3+x^2/4) - (x^4-2x^3+x^2/4). =x^3. F(y)-F(y-1) =y^3. F(z)-F(z-1) =z^3. de sorte x 3 + y ^3 =/= z ^ 3. n> 2. . Similaire. Nous avons une inférence chaîne G (z) * F (z) = G (x) * F (x) + G (y) * F (y) équivalente G (z) * F (z-1) = G (x) * F (x -1) + G (y) * F (y-1) G (z) * F (z) = G (x) * F (x) + G (y) * F (y) en déduire G (z) * F (z-1) = G (x) * F (x -1) + G (y) * F (y-1) G (z) * F (z-x-1) = G (x) * F (x-x-1) + G (y-x-1) * F (y) en déduire G (z) * F (z-x-2) = G ( x) * F (x-x-2) + G (y) * F (y-x-2) nous voyons G (z) * F (z-x-1) = G (x) * F (x-x-1) + G (y) * F (y-x-1) G (z) * F (z-x-1) = G (x) * F (-1) + G (y) * F (y-x-1) G (z) * F (z-x-1) = 0 + G (y) * F (y-x-1) donner z = y. et G (z) * F (z-x-2) = G (x) * F (x-x-2) + G (y) * F (y-x-2) G (z) * F (z-x-2) = G (x) * F (-2) + G (y) * F (y-x-2) G (z) * F (z-x-2) = G (x) + G (y) * F (y-x-2) x> 0 en déduire G (x)> 0. donner z = / = y. de sorte G (z) * F (zx-1) = G (x) * F (xx-1) + G (yx-1) * F (y) ne pas en déduire G (z) * F (z-x-2) = G (x) * F (x-x-2) + G (y) * F (y-x-2) de sorte G (z) * F (z) = G (x) * F (x) + G (y) * F (y) ne pas en déduire G (z) * F (z-1) = G (x) * F (x-1) + G (y) * F (y-1) de sorte G (z) * F (z) = G (x) * F (x) + G (y) * F (y) n'est pas équivalente G (z) * F (z-1) = G (x) * F (x-1) + G (y) * F (y-1) Donc avoir deux cas [G (x) * F (x) + G (y) * F (y)] = G (z) * F (z) et [G (x) * F (x-1) + G (y) * F (y-1)] = / = G (z-1) * F (z-1) ou vice versa. de sorte [G (x) * F (x) + G (y) * F (y)] - [G (x) * F (x-1) + G (y) * F (y-1)] = / = G (z) * [F (z)-F (z-1)]. Ou G (x) * [F (x) - F (x-1)] + G (y) * [F (y)-F (y-1)] = / = G (z) * [F (z) -F (z-1).] nous voyons x ^ n = G (x) * [F (x)-F (x-1)] y ^ n = G (y) * [F (y)-F (y-1)] z ^ n = G (z) * [F (z)-F (z-1)] de sorte x ^ n + y ^ n = / = z ^ n Le bonheur et la paix Tran Tan Cuong
