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Can u answer these for me with the work shown

Can u answer these for me with the work shown-example-1
User Neojakey
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1 Answer

4 votes

Answer:


(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

User SerjG
by
7.3k points
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