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How To Solve A Function F(x) Has These Properties The Domain Of F Is The Set Of Natural Numbers F (1)1 F (x plus 1)f(x) plus 3x(x plus 1) plus 1 A. Determine F (2) F(3) F (4) F (5) F(6) THANKS!!!!!!?
The function (sequence generator) f(x) with x∈ℕ has been defined as a recursive function (sequence), with the initial value defined for some x, ie starting form some some natural number (in this case 1), the value of the function (sequence) is given (in this case f(1) = 1), and each successive value of the function (sequence) is defined in terms of the current value f(x+1) = f{x} + g(x) where g(x) is a function with g(x) = 3x(x + 1).f(x + 1) = f(x) + 3x(x + 1)f(1) = 1→ f(2) = f(1 + 1) = f(1) + (3×1)(1 + 1) = 1 + 3×2 = 1 + 6 = 7→ f(3) = f(2 + 1) = f(2) + (3×2)(2 + 1) = 7 + 6×3 = 7 + 18 = 25I'll let you evaluate the rest.Hint:f(4) = f(3 + 1) = f(3) + (3×3)(3 + 1)f(5) = f(4 + 1) = f(4) + ...f(6) = f(5 + 1) = f(5) + ...
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 integral of f divided by the quantity 1 minus f with respect to x where f is a function of x?
∫ f(x)/(1 - f(x)) dx = -x + ∫ 1/(1 - f(x)) dx
7 plus f - 21 equals -20?
7 + f - 21 = -20 Therefore, 7 + f = 1 f = 1 - 7 f = -6
What is the reciprocal of f over b?
b/f. reciprocal of x means 1/x, i.e. 1/f/b is b/f
Why is the sum of the reciprocals of all of the divisors of a perfect number equal to 2?
Suppose N is a perfect number. Then N cannot be a square number and so N has an even number of factors.Suppose the factors are f(1) =1, f(2), f(3), ... , f(k-1), f(k)=N.Furthermore f(r) * f(k+1-r) = N for r = 1, 2, ... k so that f(r) = N/f(k+1-r)which implies that 1/f(r) = f(k+1-r)/NThen 1/f(1) + 1/(f(2) + ... + 1/f(k)= f(k)/N + f(k-1)/N + ... + f(1)/N= [f(k) + f(k-1) + ... + f(1)] / N= 2N/N since, by definition, [f(k) + f(k-1) + ... + f(1)] = 2N
F(x)=2x+3,a=-1 find 1/f. And graph f and 1/f?
Given the function f(x) = 2x + 3 and a = -1, we can find f(a) as follows: f(a) = 2(-1) + 3 f(a) = -2 + 3 f(a) = 1 So, f(a) = 1. To graph f(x) and 1/f(x), we can plot several points and connect them to visualize the functions. Here are some points for f(x): For f(x): When x = -2, f(x) = 2(-2) + 3 = -1 When x = -1, f(x) = 2(-1) + 3 = 1 When x = 0, f(x) = 2(0) + 3 = 3 When x = 1, f(x) = 2(1) + 3 = 5 When x = 2, f(x) = 2(2) + 3 = 7 Now, to find 1/f(x), we take the reciprocal of each f(x) value: For 1/f(x): When x = -2, 1/f(x) = 1/(-1) = -1 When x = -1, 1/f(x) = 1/1 = 1 When x = 0, 1/f(x) = 1/3 ≈ 0.333 When x = 1, 1/f(x) = 1/5 ≈ 0.2 When x = 2, 1/f(x) = 1/7 ≈ 0.143 Now, we can plot these points and connect them to obtain the graphs of f(x) and 1/f(x).
How To Solve A Function F(x) Has These Properties The Domain Of F Is The Set Of Natural Numbers F (1)1 F (x plus 1)f(x) plus 3x(x plus 1) plus 1 A. Determine F (2) F(3) F (4) F (5) F(6) THANKS!!!!!!?
The function (sequence generator) f(x) with x∈ℕ has been defined as a recursive function (sequence), with the initial value defined for some x, ie starting form some some natural number (in this case 1), the value of the function (sequence) is given (in this case f(1) = 1), and each successive value of the function (sequence) is defined in terms of the current value f(x+1) = f{x} + g(x) where g(x) is a function with g(x) = 3x(x + 1).f(x + 1) = f(x) + 3x(x + 1)f(1) = 1→ f(2) = f(1 + 1) = f(1) + (3×1)(1 + 1) = 1 + 3×2 = 1 + 6 = 7→ f(3) = f(2 + 1) = f(2) + (3×2)(2 + 1) = 7 + 6×3 = 7 + 18 = 25I'll let you evaluate the rest.Hint:f(4) = f(3 + 1) = f(3) + (3×3)(3 + 1)f(5) = f(4 + 1) = f(4) + ...f(6) = f(5 + 1) = f(5) + ...
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.
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.
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 3 over 6 and 1 over f?
3/6 + 1/f = 1/2 + 1/f = (f+2)/2f
How do you find the value of f(-1)?
f(-1)
What is the integral of f divided by the quantity 1 minus f with respect to x where f is a function of x?
∫ f(x)/(1 - f(x)) dx = -x + ∫ 1/(1 - f(x)) dx
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
Who is person who solve shortest Fermat?
To: trantancuong21@yahoo.com 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
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.
