Question

What is the difference of (2x+5/x^2-3x)-(3x+5/x^3-9x)-(x+1/x^2-9)

Answer 1

Answer:

The difference of given expressions is $\frac{x^2+7x+10}{x^3-9x}$.

Step-by-step explanation:

The given expression is

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

Find factors of denominators.

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

Use $a^2-b^2=(a-b)(a+b)$

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

Take  $x(x+3)(x-3)$ as LCD.

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

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

$\frac{2x^2+5x+6x+15-4x-5-x^2}{x(x^2-9)}$

$\frac{x^2+7x+10}{x^3-9x}$

Therefore the difference of given expressions is $\frac{x^2+7x+10}{x^3-9x}$.

Answer 2

Answer:

(x+5)(x+2)/x^3-9x

Step-by-step explanation:

For short if taking on edgen then option A is correct.