Answer:
1. Option A
2. Option D
Step by step explanation
1. [tex] \frac{1}{1 - x} + \frac{x}{ {x}^{2} - 1} [/tex]
Use [tex] \frac{ - a}{b} = \frac{a}{ - b} = - \frac{a}{b} [/tex] to rewrite the fractions
[tex] - \frac{1}{x - 1} + \frac{x}{(x - 1)(x + 1)} [/tex]
Write all numerators above the Least Common Denominator ( X - 1 ) ( X + 1 )
[tex] \frac{ - (x + 1) + x}{(x - 1)(x + 1)} [/tex]
When there is a ( - ) in front of an expression in parentheses , change the sign of each term in the expression
[tex] \frac{ - x - 1 + x}{(x - 1)(x + 1)} [/tex]
Using [tex](a - b)(a + b) = {a}^{2} - {b}^{2} [/tex] , simplify the product
[tex] \frac{ - x - 1 + x}{ {x}^{2} - 1 } [/tex]
Since two opposites add up to zero, remove them from the expression
[tex] \frac{ - 1}{ {x}^{2} - 1} [/tex]
So, Option A is the right option.
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2.
[tex] \frac{ {x}^{2} - x - 12}{ {x}^{2} - 16} - \frac{1 - 2x}{x + 4} [/tex]
Write - X as a difference
[tex] \frac{ {x}^{2} + 3x - 4x - 12 }{ {x}^{2} - 16 } - \frac{1 - 2x}{x + 4} [/tex]
Using [tex] {a}^{2} - {b}^{2} = (a - b)(a + b)[/tex] , factor the expression
[tex] \frac{ {x}^{2} + 3x - 4x - 12 }{(x - 4)(x + 4)} - \frac{1 - 2x}{x + 4} [/tex]
Factor the expression
[tex] \frac{x(x + 3) - 4(x + 3)}{(x - 4)(x + 4)} - \frac{1 - 2x}{x + 4} [/tex]
Factor out X+3 from the expression
[tex] \frac{(x + 3)(x - 4)}{(x - 4)(x + 4)} - \frac{1 - 2x}{x + 4} [/tex]
Reduce the fraction with x-4
[tex] \frac{x + 3}{x + 4} - \frac{1 - 2x}{x + 4} [/tex]
Write all the numerators above the common denominator
[tex] \frac{x + 3 - ( 1- 2x)}{x + 4} [/tex]
When there is a (-) in front of an expression in parentheses, change the sign of each term in the expression
[tex] \frac{x + 3 - 1 + 2x}{x + 4} [/tex]
Collect like terms
[tex] \frac{3x + 3 - 1}{x + 4} [/tex]
Subtract the numbers
[tex] \frac{3x + 2}{x + 4} [/tex]
Undefined at,
X + 4 = 0
Move constant to R.H.S and change its sign
[tex]x = 0 - 4[/tex]
Calculate
[tex]x = - 4[/tex]
So, the answer is :
[tex] \frac{3x + 2}{x + 4} [/tex] , undefined at X = -4 and 4
Hope this helps..
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