Đáp án:
a) \(\frac{4x}{2x+3}\)
b)\(\frac{16}{5}\)
Giải thích các bước giải:
a)T=\((\frac{4x^{2}}{x^{2}-4}-\frac{2+x}{2-x}+\frac{2-x}{x+2}):\frac{2x+3}{x-2}\)
= \((\frac{4x^{2}}{x^{2}-4}+\frac{x+2}{x-2}+\frac{2-x}{x+2}):\frac{2x+3}{x-2}\)
=\(\frac{4x^{2}+x^{2}+4x+4-x^{2}+4x-4}{(x+2)\cdot (x-2)}:\frac{2x+3}{x-2}\)
=\(\frac{4x^{2}+8x}{(x+2)\cdot (x-2)}:\frac{2x+3}{x-2}\)
=\(\frac{4x\cdot (x+2)}{(x+2)\cdot (x-2)}\cdot \frac{x-2}{2x+3}\)
=\(\frac{4x}{2x+3}\)
b) Khi x=\(\frac{\frac{-2}{1}}{2}=-4\)
⇒ T=\(\frac{4x}{2x+3}=\frac{4\cdot (-4)}{2\cdot (-4)+3}=\frac{16}{5}\)
