Question

plz solve this problem​

Answer 1

Given:

$\dfrac{6x^2-7x-5}{8+2x-x^2}+\dfrac{9x^2-12x-5}{4+11x-3x^2}$

To find:

Simplified form of given expression.

Solution:

We have,

$\dfrac{6x^2-7x-5}{8+2x-x^2}+\dfrac{9x^2-12x-5}{4+11x-3x^2}$

$=\dfrac{6x^2-10x+3x-5}{8+4x-2x-x^2}+\dfrac{9x^2-15x+3x-5}{4+12x-x-3x^2}$

$=\dfrac{2x(3x-5)+(3x-5)}{4(2+x)-x(2+x)}+\dfrac{3x(3x-5)+(3x-5)}{4(1+3x)-x(1+3x)}$

$=\dfrac{(3x-5)(2x+1)}{(2+x)(4-x)}+\dfrac{3x-5}{4-x}$

$=\dfrac{(3x-5)(2x+1)+(3x-5)(2+x)}{(2+x)(4-x)}$

$=\dfrac{(3x-5)(2x+1+2+x)}{(2+x)(4-x)}$

$=\dfrac{(3x-5)(3x+3)}{(2+x)(4-x)}$

Therefore, the simplified form of given expression is $\dfrac{(3x-5)(3x+3)}{(2+x)(4-x)}$.