Question

Can u answer these for me with the work shown

Answer 1

`(x^3 + 2x^2 -9x-18)/(x^3-x^2-6x)= ((x+3))/(x)`

`(3x^2 - 5x - 2)/(x^3 - 2x^2) = (3x + 1)/(x^2)`

`(6 - 2x)/(x^2 - 9) * (15 + 5x)/(4x)=-(5)/(2x)`

`(x^2 -6x + 9)/(5x - 15) / (5)/(3-x) = (-(x-3)^2)/(25)`

`(x^3 - x^2 -x + 1)/(x^2 - 2x+1)= x +1`

`(9x^2 + 3x)/(6x^2) = (3x + 1)/(2x)`

`(x^2-3x+2)/(4x) * (12x^2)/(x^2 - 2x) / (x - 1)/(x) = 3x`

Explanation:

Required

Simplify

Solving (1):

`(x^3 + 2x^2 -9x-18)/(x^3-x^2-6x)`

Factorize the numerator and the denominator

`(x^2(x + 2) -9(x+2))/(x(x^2-x-6))`

Factor out x+2 at the numerator

`((x^2 -9)(x+2))/(x(x^2-x-6))`

Express x^2 - 9 as difference of two squares

`((x^2 -3^2)(x+2))/(x(x^2-x-6))`

`((x -3)(x+3)(x+2))/(x(x^2-x-6))`

Expand the denominator

`((x -3)(x+3)(x+2))/(x(x^2-3x+2x-6))`

Factorize

`((x -3)(x+3)(x+2))/(x(x(x-3)+2(x-3)))`

`((x -3)(x+3)(x+2))/(x(x+2)(x-3))`

Cancel out same factors

`((x+3))/(x)`

Hence:

`(x^3 + 2x^2 -9x-18)/(x^3-x^2-6x)= ((x+3))/(x)`

Solving (2):

`(3x^2 - 5x - 2)/(x^3 - 2x^2)`

Expand the numerator and factorize the denominator

`(3x^2 - 6x + x - 2)/(x^2(x- 2))`

Factorize the numerator

`(3x(x - 2) + 1(x - 2))/(x^2(x- 2))`

Factor out x - 2

`((3x + 1)(x - 2))/(x^2(x- 2))`

Cancel out x - 2

`(3x + 1)/(x^2)`

Hence:

`(3x^2 - 5x - 2)/(x^3 - 2x^2) = (3x + 1)/(x^2)`

Solving (3):

`(6 - 2x)/(x^2 - 9) * (15 + 5x)/(4x)`

Express x^2 - 9 as difference of two squares

`(6 - 2x)/(x^2 - 3^2) * (15 + 5x)/(4x)`

Factorize all:

`(2(3 - x))/((x- 3)(x+3)) * (5(3 + x))/(2(2x))`

Cancel out x + 3 and 3 + x

`(2(3 - x))/((x- 3)) * (5)/(2(2x))`

`(3 - x)/(x- 3) * (5)/(2x)`

Express
`3 - x` as
`-(x - 3)`

`(-(x-3))/(x- 3) * (5)/(2x)\\`

`-1 * (5)/(2x)`

`-(5)/(2x)`

Hence:

`(6 - 2x)/(x^2 - 9) * (15 + 5x)/(4x)=-(5)/(2x)`

Solving (4):

`(x^2 -6x + 9)/(5x - 15) / (5)/(3-x)`

Expand x^2 - 6x + 9 and factorize 5x - 15

`(x^2 -3x -3x+ 9)/(5(x - 3)) / (5)/(3-x)`

Factorize

`(x(x -3) -3(x-3))/(5(x - 3)) / (5)/(3-x)`

`((x -3)(x-3))/(5(x - 3)) / (5)/(3-x)`

Cancel out x - 3

`((x -3))/(5) / (5)/(3-x)`

Change / to *

`((x -3))/(5) * (3-x)/(5)`

Express
`3 - x` as
`-(x - 3)`

`((x -3))/(5) * (-(x-3))/(5)`

`(-(x-3)(x -3))/(5*5)`

`(-(x-3)^2)/(25)`

Hence:

`(x^2 -6x + 9)/(5x - 15) / (5)/(3-x) = (-(x-3)^2)/(25)`

Solving (5):

`(x^3 - x^2 -x + 1)/(x^2 - 2x+1)`

Factorize the numerator and expand the denominator

`(x^2(x - 1) -1(x - 1))/(x^2 - x-x+1)`

Factor out x - 1 at the numerator and factorize the denominator

`((x^2 - 1)(x - 1))/(x(x -1)- 1(x-1))`

Express x^2 - 1 as difference of two squares and factor out x - 1 at the denominator

`((x +1)(x-1)(x - 1))/((x -1)(x-1))`

`x +1`

Hence:

`(x^3 - x^2 -x + 1)/(x^2 - 2x+1)= x +1`

Solving (6):

`(9x^2 + 3x)/(6x^2)`

Factorize:

`(3x(3x + 1))/(3x(2x))`

Divide by 3x

`(3x + 1)/(2x)`

Hence:

`(9x^2 + 3x)/(6x^2) = (3x + 1)/(2x)`

Solving (7):

`(x^2-3x+2)/(4x) * (12x^2)/(x^2 - 2x) / (x - 1)/(x)`

Change / to *

`(x^2-3x+2)/(4x) * (12x^2)/(x^2 - 2x) * (x)/(x-1)`

Expand

`(x^2-2x-x+2)/(4x) * (12x^2)/(x^2 - 2x) * (x)/(x-1)`

Factorize

`(x(x-2)-1(x-2))/(4x) * (12x^2)/(x(x - 2)) * (x)/(x-1)`

`((x-1)(x-2))/(4x) * (12x^2)/(x(x - 2)) * (x)/(x-1)`

Cancel out x - 2 and x - 1

`(1)/(4x) * (12x^2)/(x) * (x)/(1)`

Cancel out x

`(1)/(4x) * (12x^2)/(1) * (1)/(1)`

`(12x^2)/(4x)`

`3x`

Hence:

`(x^2-3x+2)/(4x) * (12x^2)/(x^2 - 2x) / (x - 1)/(x) = 3x`