Giải Bài tập 18 trang 41 sgk Toán 8 tập 1 Chân trời

a) $\frac{2x^{2}-1}{x-2}+\frac{-x^{2}-3}{x-2}=\frac{2x^{2}-1-x^{2}-3}{x-2}=\frac{x^{2}-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2$

b) $\frac{x}{x+y}+\frac{y}{x-y}=\frac{x^{2}-xy+xy+y^{2}}{(x-y)(x+y)}=\frac{x^{2}+y^{2}}{x^{2}-y^{2}}$

c) $\frac{1}{x-1}-\frac{2}{x^{2}-1}=\frac{x+1-2}{(x-1)(x+1)}=\frac{x-1}{(x-1)(x+1)}=\frac{1}{x+1}$

d) $\frac{x+2}{x^{2}+xy}-\frac{y-2}{xy+y^{2}}=\frac{xy+2y-xy+2x}{xy(x+y)}=\frac{2(x+y)}{xy(x+y)}=\frac{2}{xy}$

e) $\frac{1}{2x^{2}-3x}-\frac{1}{4x^{2}-9}=\frac{2x+3-x}{x(2x-3)(2x+3)}=\frac{x+3}{x(2x-3)(2x+3)}$

g) $\frac{2x}{9-x^{2}}+\frac{1}{x-3}-\frac{1}{x+3}=\frac{-2x+x+3-x+3}{(x+3)(x-3)}=\frac{-2x+6}{(x+3)(x-3)}=\frac{-2(x-3)}{(x-3)(x+3)}=\frac{-2}{x+3}$