Question

What is the difference?

Answer 1

Answer:

The option $\frac{(x+5)(x+2)}{x^3-9x}$ is correct

The difference of the given expression is

$\frac{2x+5}{x^2-3x}-(\frac{3x+5}{x^3-9x})-({\frac{x+1}{x^2-9})=\frac{(x+5)(x+2)}{x^3-9x}$

Step-by-step explanation:

Given expression is $\frac{2x+5}{x^2-3x}-(\frac{3x+5}{x^3-9x})-({\frac{x+1}{x^2-9})$

To find the difference of the given expression as below :

$\frac{2x+5}{x^2-3x}-(\frac{3x+5}{x^3-9x})-({\frac{x+1}{x^2-9})$

$=\frac{2x+5}{x(x-3)}-(\frac{3x+5}{x(x^2-9)})-({\frac{x+1}{x^2-9})$

$=\frac{2x+5}{x(x-3)}-(\frac{3x+5}{x(x^2-3^2)})-({\frac{x+1}{x^2-3^2})$

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

( using the formula $a^2-b^2=(a+b)(a-b)$ )

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

$=\frac{2x^2+6x+5x+15-3x-5-x^2-x}{x(x-3)(x+3)}$ (adding the like terms)

$=\frac{x^2+7x+10}{x(x^2-9)}$ ( by factoring the quadratic polynomial )

$=\frac{(x+5)(x+2)}{x^3-9x}$

Therefore $\frac{2x+5}{x^2-3x}-(\frac{3x+5}{x^3-9x})-({\frac{x+1}{x^2-9})=\frac{(x+5)(x+2)}{x^3-9x}$

Therefore the difference of the given expression is

$\frac{(x+5)(x+2)}{x^3-9x}$

Therefore option $\frac{(x+5)(x+2)}{x^3-9x}$ is correct