Giải bài tập 6.23 trang 19 Toán 8 tập 2 KNTT

a) $\frac{x^{2}+4x+4}{x^{2}-4}+\frac{x}{2-x}+\frac{4-x}{5x-10}$

$=\frac{(x+2)^{2}}{(x-2)(x+2)}-\frac{x}{x-2}+\frac{4-x}{5(x-2)}$

$=\frac{x+2}{x-2}-\frac{x}{x-2}+\frac{4-x}{5(x-2)}$

$=\frac{5(x+2)-5x+4-x}{5(x-2)}$

$=\frac{-x+14}{5(x-2)}$

b) $\frac{x}{x^{2}+1}-\left ( \frac{3}{x+6}+\frac{x-2}{x+4} \right )+\left [ \frac{3}{x+6}-\left ( \frac{1}{x^{2}+1}-\frac{x-2}{x+4} \right ) \right ]$

$=\frac{x}{x^{2}+1}-\frac{3}{x+6}-\frac{x-2}{x+4}+\frac{3}{x+6}-\frac{1}{x^{2}+1}+\frac{x-2}{x+4}$

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

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