Question

Bài 1 :Tính 

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

Answer 1

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

\(=\frac{x+4}{\left(x-2\right)\left(x+2\right)}-\frac{2}{x\left(x+2\right)}\)

\(=\frac{\left(x+4\right).x}{x\left(x-2\right)\left(x+2\right)}-\frac{2\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2+4x-2x-4}{x\left(x-2\right)\left(x+2\right)}=\frac{x^2+2x-4}{x\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2-2^2+2x}{x\left(x-2\right)\left(x+2\right)}=\frac{\left(x+2\right)\left(x-2\right)+2x}{x\left(x-2\right)\left(x+2\right)}=\frac{2x}{x}\)

Answer 2

\(\frac{x+4}{x^2-4}-\frac{2}{x^2+2x}ĐK:x\ne\pm2;0\)

\(=\frac{x+4}{\left(x-2\right)\left(x+2\right)}-\frac{2}{x\left(x+2\right)}=\frac{x\left(x+4\right)}{x\left(x-2\right)\left(x+2\right)}-\frac{2\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2+4x-2x+4}{x\left(x-2\right)\left(x+2\right)}=\frac{x^2-2x+4}{x\left(x-2\right)\left(x+2\right)}\)

\(=\frac{\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)}=\frac{x-2}{x\left(x+2\right)}\)

Answer 3

Bạn làm +4x-2x sao lại ra -2x vậy???