Question

2x/(x²-4) + 1/(x²+3x+2)
solve this with steps​

Answer 1

To solve :-

$\tt \implies \dfrac{2x}{ {x}^{2} - 4 } + \dfrac{1}{ {x}^{2} + 3x + 2 }$

Solution :-

$\tt \implies \dfrac{2x}{ {x}^{2} - 4 } + \dfrac{1}{ {x}^{2} + 3x + 2 }$

$\tt \implies \dfrac{2x}{ {x}^{2} - {2}^{2} } + \dfrac{1}{ {x}^{2} + 3x + 2 }$

Factorise denominator of 2nd term

$\tt \implies \dfrac{2x}{ {x}^{2} - {2}^{2} } + \dfrac{1}{ {x}^{2} + 2x + x + 2 }$

$\tt \implies \dfrac{2x}{ {x}^{2} - {2}^{2} } + \dfrac{1}{ x(x + 2)+ 1( x + 2) }$

$\tt \implies \dfrac{2x}{ {x}^{2} - {2}^{2} } + \dfrac{1}{ (x + 2)( x + 1) }$

Apply formula (a²-b²) = (a+b)(a-b) in denominator of first term where a = x and b =2

$\tt \implies \dfrac{2x}{(x + 2)(x - 2) } + \dfrac{1}{ (x + 2)( x + 1) }$

Now take LCM

$\tt \implies \dfrac{2x(x + 1) + 1(x - 2)}{(x + 2)(x - 2)( x + 1) }$

$\tt \implies \dfrac{2 {x}^{2} + 2x+ x - 2}{(x + 2)(x - 2)( x + 1) }$

$\tt \implies \dfrac{2 {x}^{2} + 3 x - 2}{(x + 2)(x - 2)( x + 1) }$

Now factorise numerator

$\tt \implies \dfrac{2 {x}^{2} + 4x - x - 2}{(x + 2)(x - 2)( x + 1) }$

$\tt \implies \dfrac{2 x(x + 2) - 1(x + 2)}{(x + 2)(x - 2)( x + 1) }$

$\tt \implies \dfrac{(2x - 1)(x + 2)}{(x + 2)(x - 2)( x + 1) }$

Cancel out like terms from numerator and denominator

$\tt \implies \dfrac{(2x - 1) \cancel{(x + 2)}}{ \cancel{(x + 2)}(x - 2)( x + 1) }$

$\tt \implies \dfrac{(2x - 1)}{ (x - 2)( x + 1) }$

$\tt \implies \dfrac{(2x - 1)}{ x ( x + 1) - 2(x + 1) }$

$\tt \implies \dfrac{(2x - 1)}{ {x}^{2} + x - 2x - 2 }$

$\tt \implies \dfrac{2x - 1}{ {x}^{2} - x - 2 }$

Since this can't be solved further, it's your final answer.