Question

rút gọn :  \(\left(\frac{x}{x^2-16}-\frac{x-4}{x^2+4x}\right):\frac{2x-4}{x^2+4x}\)

Answer 1

\(\left[\frac{x}{\left(x+4\right)\left(x-4\right)}-\frac{x-4}{x\left(x+4\right)}\right]:\frac{2\left(x-2\right)}{x\left(x+4\right)}\)\(=\left[\frac{x^2-\left(x-4\right)^2}{x\left(x+4\right)\left(x-4\right)}\right].\left[\frac{x\left(x+4\right)}{2\left(x-2\right)}\right]\)\(=\left(\frac{x^2-x^2+8x-16}{x\left(x+4\right)\left(X-4\right)}\right).\frac{x\left(x+4\right)}{2\left(x-2\right)}=\frac{8\left(x-2\right).x\left(x+4\right)}{x\left(x+4\right)\left(x-4\right).2\left(x-2\right)}=\frac{4}{x-4}\)