Đáp án:a)\(\frac{2}{x}\)
b)x-1
Giải thích các bước giải:
a)\(\frac{3}{x+3}-\frac{x-6}{x^{2}+3x}=\frac{3}{x+3}-\frac{x-6}{x(x+3)}=\frac{3x-x+6}{x(x+3)}=\frac{2x+6}{x(x+3)}\)
=\(\frac{2(x+3)}{x(x+3)}=\frac{2}{x}\)
b)đk: \(x\neq1\)
\(\frac{2x^{2}-x}{x-1}+\frac{x+1}{1-x}+\frac{2-x^{2}}{x-1}\)
=\(\frac{2x^{2}-x}{x-1}-\frac{x-1}{x-1}+\frac{2-x^{2}}{x-1}\)
=\(\frac{x^{2}-2x+1}{x-1}=\frac{(x-1)^{2}}{x-1}=x-1\)