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Simplify by dividing x3 plus 2x2 plus 3x-6 over x-1?
(x^3 + 2x^2 + 3x - 6)/(x - 1) add and subtract x^2, and write -6 as (- 3) + (-3) = (x^3 - x^2 + x^2 + 2x^2 - 3 + 3x - 3)/(x - 1) = [(x^3 - x^2) + (3x^2 - 3) + (3x - 3)]/(x - 1) = [x^2(x - 1) + 3(x^2 - 1) + 3(x - 1)]/(x - 1) = [x^2(x - 1) + 3(x - 1)(x + 1) + 3(x - 1)]/(x - 1) = [(x - 1)(x^2 + 3x + 3 + 3)]/(x - 1) = x^2 + 3x + 6
What is 54.99 plus 3 x 1 plus 4 x 1 plus 3 x 3 plus 4 x 1 plus 3 x 1 plus 3 x 2 plus 3 x 3 plus 4 x 1 plus 3 x 1 x 3 x 3?
It is 123.99
Who do solve FLT some lines?
I like back into multidimensional space where were born to be sleeping forever thousands autumn. Pierre De Fermat's last theorem. The conditions. x,y,z,n are the integers >0 and n>2. z^n=/x^n+y^n. Assumptions z^3=x^3+y^3. Therefore z=(x^3+y^3)^1/3. I define . F(x,y)=(x^3+y^3}^1/3 - [ (x-x-1)^3+(y-x-1)^3]^1/3. Therefore [z-F(x,y)]^3={ (x^3+y^3)^1/3 - (x^3+y^3}^1/3 + [ (x-x-1)^3+(y-x-1)^3] ^1/3 }^3={ [(x-x-1)^3+(y-x-1)^3]^1/3 }^3 =(x-x-1)^3+(y-x-1)^3= (y-x-1)^3-1. Because [z-F(x,y)]^3=(y-x-1)^3-1 . Attention [(y-x-1)^3-1 ] is an integer , [(y-x-1)^3-1 ]^1/3 is an irrational number therefore [(y-x-1)^3-1 ]^2/3 is an irrational number too. Example (2^3-1) is an integer , (2^3-1)^1/3 is an irrational number and (2^3-1)^2/3 is an irrational number too. Because z-F(x,y)=[y-x-1)^3-1]^1/3. Therefore z=F(x,y)+[(y-x-1)^3-1] ^1/3. Therefore z^3=[F(x,y)]^3.+3[F(x,y)]^2*[(y-x-1)^3-1]^1/3+3F(x,y)*[(y-x-1)-1]^2/3+[(y-x-1)^3-1]. Therefore 3F(x,y)*[(y-x-1)^3-1]^2/3+[F(x,y)]^3+3[F(x,y)]^2*[(y-x-1)^3-1]^1/3+[(y-x-1)^3-1] - z^3 =0 Because. z-F(x,y)=[ (y-x-1)^3-1]^1/3. Therefore F(x,y)=z - [(y-x-1)^3-1]^1/3. Therefore 3F(x,y)*[(y-x-1)^3-1]^2/3 =3z*[(y-x-1)^3-1]^2/3 -3[(y-x-1)^3-1]. Therefore. 3z*[(y-x-1)^3-1]^2/3 - 3[(y-x-1)^3-1]+[F(x,y)]^3+3[F(x,y)]^2*[y-x-1)^3-1]^1/3+[(y-x-1)^3-1] - z^3=0.. Named [F(x,y)]^3+3[F(x,y)]^2*[(y-x-1)^3-1]^1/3+[(y-x-1)^3-1] - z^3= W. We have 3z*[(y-x-1)^3-1]^2/3 is an irrational number because z is an integer and had proved [(y-x-1)^3-1]^2/3 is an irrational number. And 3[(y-x-1)^3-1] is an integer because x,y are the integers. And 3z*[(y-x-1)^3-1]^2/3 - 3[(y-x-1)^3-1]+[F(x,y)]^3+W=0. Therefore an irrational number - an integer+W=0. Therefore W is an complex irrational number. Named 3z*[(y-x-1)^3-1]^2/3 is 3z*B And Named 3[(y-x-1)^3-1] is C . Therefore 3z*B - C +W=0. Therefore 3z*B=C-W. Because z is an integer. B is an irrational number=[(y-x-1)^3-1]^2/3 Attention (an integer)^2/3 and (an integer)^2/3 is an irrational number. W is an complex irrational number. C is an integer. Therefore. an integer*an irrational number=an complex irrational number + an integer. Unreasonable. Therefore. z^3=/x^3+y^3 Similar z^n=/x^n+y^n. ISHTAR.
What is the third to the fourth power?
1/3 x 1/3 x 1/3 x 1/3 =1/81
What is 1 over x plus 3 plus 1 divided by 1 over x-3?
There is some ambiguity in the question because of the absence of brackets. For example, is it 1/x + 3 ... or 1/(x+3) ... ? So, this is my interpretation of the question. If it is not what you meant please resubmit using brackets. 1/(x+3) + 1/1/(x-3) = 1/(x+3) + (x - 3) = (1 + x2 - 9)/(x + 3) = (x2 - 8)/(x + 3)
Factor x3 1?
x^3 + 1 = (x + 1)(x^2 - x + 1) x^3 - 1 = (x - 1)(x^2 + x + 1)
What is one third percent of 1500?
1/3% = (1/3) x 1% = 1/3 x 0.011/3 x 0.01 x 1,500 = 1/3 x 15 = 5
What is x2 plus 3x plus x plus 3?
X2+3x+x+3=x(x+1)+3(x+1)=(x+3)(x+1)
What 3 numbers multiplies gives you 27?
3³ = 3 x 3 x 3 = 27 1 x 3 x 9 = 27 1 x 1 x 27 = 27
What is the derivative of x divided by 3?
The derivative of x divided by 3 is 1/3. This can be found using the power rule of differentiation, where the derivative of x^n is nx^(n-1). In this case, x can be written as x^1, so the derivative is 1(1/3)*x^(1-1) = 1/3.
What is 3 multiplied 3 and one half?
3 x 3 x 1/23/1 x 3/1 x 1/2 = 9/2 = 4.5
Solve the inequality x3 2x2-3x0?
If that's + 2x2, the answer is x(x + 3)(x - 1) x = 0, -3, 1 If that's - 2x2, the answer is x(x - 3)(x + 1) x = 0, 3, -1
