Answer:
[tex]\frac{x^3 + 2x^2 -9x-18}{x^3-x^2-6x}= \frac{(x+3)}{x}[/tex]
[tex]\frac{3x^2 - 5x - 2}{x^3 - 2x^2} = \frac{3x + 1}{x^2}[/tex]
[tex]\frac{6 - 2x}{x^2 - 9} * \frac{15 + 5x}{4x}=-\frac{5}{2x}[/tex]
[tex]\frac{x^2 -6x + 9}{5x - 15} / \frac{5}{3-x} = \frac{-(x-3)^2}{25}[/tex]
[tex]\frac{x^3 - x^2 -x + 1}{x^2 - 2x+1}= x +1[/tex]
[tex]\frac{9x^2 + 3x}{6x^2} = \frac{3x + 1}{2x}[/tex]
[tex]\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} / \frac{x - 1}{x} = 3x[/tex]
Step-by-step explanation:
Required
Simplify
Solving (1):
[tex]\frac{x^3 + 2x^2 -9x-18}{x^3-x^2-6x}[/tex]
Factorize the numerator and the denominator
[tex]\frac{x^2(x + 2) -9(x+2)}{x(x^2-x-6)}[/tex]
Factor out x+2 at the numerator
[tex]\frac{(x^2 -9)(x+2)}{x(x^2-x-6)}[/tex]
Express x^2 - 9 as difference of two squares
[tex]\frac{(x^2 -3^2)(x+2)}{x(x^2-x-6)}[/tex]
[tex]\frac{(x -3)(x+3)(x+2)}{x(x^2-x-6)}[/tex]
Expand the denominator
[tex]\frac{(x -3)(x+3)(x+2)}{x(x^2-3x+2x-6)}[/tex]
Factorize
[tex]\frac{(x -3)(x+3)(x+2)}{x(x(x-3)+2(x-3))}[/tex]
[tex]\frac{(x -3)(x+3)(x+2)}{x(x+2)(x-3)}[/tex]
Cancel out same factors
[tex]\frac{(x+3)}{x}[/tex]
Hence:
[tex]\frac{x^3 + 2x^2 -9x-18}{x^3-x^2-6x}= \frac{(x+3)}{x}[/tex]
Solving (2):
[tex]\frac{3x^2 - 5x - 2}{x^3 - 2x^2}[/tex]
Expand the numerator and factorize the denominator
[tex]\frac{3x^2 - 6x + x - 2}{x^2(x- 2)}[/tex]
Factorize the numerator
[tex]\frac{3x(x - 2) + 1(x - 2)}{x^2(x- 2)}[/tex]
Factor out x - 2
[tex]\frac{(3x + 1)(x - 2)}{x^2(x- 2)}[/tex]
Cancel out x - 2
[tex]\frac{3x + 1}{x^2}[/tex]
Hence:
[tex]\frac{3x^2 - 5x - 2}{x^3 - 2x^2} = \frac{3x + 1}{x^2}[/tex]
Solving (3):
[tex]\frac{6 - 2x}{x^2 - 9} * \frac{15 + 5x}{4x}[/tex]
Express x^2 - 9 as difference of two squares
[tex]\frac{6 - 2x}{x^2 - 3^2} * \frac{15 + 5x}{4x}[/tex]
Factorize all:
[tex]\frac{2(3 - x)}{(x- 3)(x+3)} * \frac{5(3 + x)}{2(2x)}[/tex]
Cancel out x + 3 and 3 + x
[tex]\frac{2(3 - x)}{(x- 3)} * \frac{5}{2(2x)}[/tex]
[tex]\frac{3 - x}{x- 3} * \frac{5}{2x}[/tex]
Express [tex]3 - x[/tex] as [tex]-(x - 3)[/tex]
[tex]\frac{-(x-3)}{x- 3} * \frac{5}{2x}\\[/tex]
[tex]-1 * \frac{5}{2x}[/tex]
[tex]-\frac{5}{2x}[/tex]
Hence:
[tex]\frac{6 - 2x}{x^2 - 9} * \frac{15 + 5x}{4x}=-\frac{5}{2x}[/tex]
Solving (4):
[tex]\frac{x^2 -6x + 9}{5x - 15} / \frac{5}{3-x}[/tex]
Expand x^2 - 6x + 9 and factorize 5x - 15
[tex]\frac{x^2 -3x -3x+ 9}{5(x - 3)} / \frac{5}{3-x}[/tex]
Factorize
[tex]\frac{x(x -3) -3(x-3)}{5(x - 3)} / \frac{5}{3-x}[/tex]
[tex]\frac{(x -3)(x-3)}{5(x - 3)} / \frac{5}{3-x}[/tex]
Cancel out x - 3
[tex]\frac{(x -3)}{5} / \frac{5}{3-x}[/tex]
Change / to *
[tex]\frac{(x -3)}{5} * \frac{3-x}{5}[/tex]
Express [tex]3 - x[/tex] as [tex]-(x - 3)[/tex]
[tex]\frac{(x -3)}{5} * \frac{-(x-3)}{5}[/tex]
[tex]\frac{-(x-3)(x -3)}{5*5}[/tex]
[tex]\frac{-(x-3)^2}{25}[/tex]
Hence:
[tex]\frac{x^2 -6x + 9}{5x - 15} / \frac{5}{3-x} = \frac{-(x-3)^2}{25}[/tex]
Solving (5):
[tex]\frac{x^3 - x^2 -x + 1}{x^2 - 2x+1}[/tex]
Factorize the numerator and expand the denominator
[tex]\frac{x^2(x - 1) -1(x - 1)}{x^2 - x-x+1}[/tex]
Factor out x - 1 at the numerator and factorize the denominator
[tex]\frac{(x^2 - 1)(x - 1)}{x(x -1)- 1(x-1)}[/tex]
Express x^2 - 1 as difference of two squares and factor out x - 1 at the denominator
[tex]\frac{(x +1)(x-1)(x - 1)}{(x -1)(x-1)}[/tex]
[tex]x +1[/tex]
Hence:
[tex]\frac{x^3 - x^2 -x + 1}{x^2 - 2x+1}= x +1[/tex]
Solving (6):
[tex]\frac{9x^2 + 3x}{6x^2}[/tex]
Factorize:
[tex]\frac{3x(3x + 1)}{3x(2x)}[/tex]
Divide by 3x
[tex]\frac{3x + 1}{2x}[/tex]
Hence:
[tex]\frac{9x^2 + 3x}{6x^2} = \frac{3x + 1}{2x}[/tex]
Solving (7):
[tex]\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} / \frac{x - 1}{x}[/tex]
Change / to *
[tex]\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} * \frac{x}{x-1}[/tex]
Expand
[tex]\frac{x^2-2x-x+2}{4x} * \frac{12x^2}{x^2 - 2x} * \frac{x}{x-1}[/tex]
Factorize
[tex]\frac{x(x-2)-1(x-2)}{4x} * \frac{12x^2}{x(x - 2)} * \frac{x}{x-1}[/tex]
[tex]\frac{(x-1)(x-2)}{4x} * \frac{12x^2}{x(x - 2)} * \frac{x}{x-1}[/tex]
Cancel out x - 2 and x - 1
[tex]\frac{1}{4x} * \frac{12x^2}{x} * \frac{x}{1}[/tex]
Cancel out x
[tex]\frac{1}{4x} * \frac{12x^2}{1} * \frac{1}{1}[/tex]
[tex]\frac{12x^2}{4x}[/tex]
[tex]3x[/tex]
Hence:
[tex]\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} / \frac{x - 1}{x} = 3x[/tex]
