\(\frac{3}{a+b}\)
\(\frac{a-3b}{a^2-b^2}\)
\(\frac{3a-b}{a^2 – b^2}\)
\(\frac{a-3b}{a^2+b^2}\)
Correct answer is B
Simplify \(\frac{2}{a+b}-\frac{1}{a-b}; \frac{2(a-b)-1(a+b)}{(a+b)(a-b)}\)
= \(\frac{2a-2b-a-b}{(a+b)(a-b)}\)
= \(\frac{a-3b}{a^2 - ab + ab - b^2}\)
= \(\frac{a-3b}{a^2-b^2}\)