Answer:
Option b) is correct.
The difference of given expression is
[tex]\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)}[/tex]
Step-by-step explanation:
Given expression is
[tex]\left(\frac{2x+5}{x^2-3x}\right)-\left(\frac{3x+5}{x^3-9x}\right)-\left(\frac{1x+1}{x^2-9}\right)[/tex]
To find their difference
[tex]\left(\frac{2x+5}{x^2-3x}\right)-\left(\frac{3x+5}{x^3-9x}\right)-\left(\frac{x+1}{x^2-9}\right)[/tex]
The expression can be written as below
[tex]\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)[/tex]
[tex]=\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)[/tex]
[tex]=\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)[/tex] (using [tex]a^2-b^2=(a+b)(a-b)[/tex])
[tex]=\frac{(2x+5)(x+3)-(3x+5)-(x+1)x}{x(x+3)(x-3)}[/tex]
[tex]=\frac{2x^2+6x+5x+15-3x-5-x^2-x}{x(x+3)(x-3)}[/tex]
[tex]=\frac{x^2+7x+10}{x(x+3)(x-3)}[/tex]
[tex]=\frac{(x+2)(x+5)}{x(x+3)(x-3)}[/tex]
[tex]=\frac{(x+2)(x+5)}{x(x^2-3^2)}[/tex] (using [tex]a^2-b^2=(a+b)(a-b)[/tex])
[tex]=\frac{(x+2)(x+5)}{x(x^2-9)}[/tex]
[tex]=\frac{(x+2)(x+5)}{(x^3-9x)}[/tex]
Therefore [tex]\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)}[/tex]
Option b) is correct.
The difference of given expression is
[tex]\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)}[/tex]
