Question

What is the difference?

Answer 1

Answer:

Option b) is correct.

The difference of given expression is

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

Step-by-step explanation:

Given expression is

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

To find their difference

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

The expression can be written as below

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

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

$=\left(\frac{2x+5}{x(x-3)}\right)-\left(\frac{3x+5}{x(x+3)(x-3)}\right)-\left(\frac{x+1}{(x+3)(x-3)}\right)$ (using $a^2-b^2=(a+b)(a-b)$)

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

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

$=\frac{x^2+7x+10}{x(x+3)(x-3)}$

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

$=\frac{(x+2)(x+5)}{x(x^2-3^2)}$   (using $a^2-b^2=(a+b)(a-b)$)

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

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

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

Option b) is correct.

The difference of given expression is

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