Question

A= ( \(\dfrac{4x}{2+x}\)+ \(\dfrac{8x^2}{4-x^2}\)) : ( \(\dfrac{x-1}{x^{2^{ }}-2x}\) - \(\dfrac{2}{x}\))

a, Rút gọn A.

b, Tìm x biết A>0

Answer 1

Lời giải:

ĐK: \(x\neq \pm 2;x\neq 0\)

a) Ta có:

\(A=\left(\frac{4x}{2+x}+\frac{8x^2}{(2-x)(2+x)}\right):\left(\frac{x-1}{x(x-2)}-\frac{2}{x}\right)\)

\(=\frac{4x(2-x)+8x^2}{(2-x)(2+x)}:\frac{(x-1)-2(x-2)}{x(x-2)}\)

\(=\frac{8x+4x^2}{(2-x)(2+x)}:\frac{-x+3}{x(x-2)}\)

\(=\frac{4x(x+2)}{(2-x)(2+x)}.\frac{x(x-2)}{3-x}\)

\(=\frac{4x}{2-x}.\frac{x(2-x)}{x-3}=\frac{4x^2}{x-3}\)

b) Để \(A>0\) thì \(\frac{4x^2}{x-3}>0\)

Mà \(4x^2>0, \forall x\neq 0\), do đó để \(\frac{4x^2}{x-3}>0\Rightarrow x-3>0\Rightarrow x>3\)

Vậy $x>3$