Câu hỏi của Duong Thi Nhuong

Question

Rút gọn : $\frac{2+x}{2-x}\div\frac{4x^2}{4-4x+x^2}.\left(\frac{2}{2-x}-\frac{8}{8+x^3}.\frac{4-2x+x^2}{2-x}\right)$

Phương An · Accepted Answer

$\frac{2+x}{2-x}\div\frac{4x^2}{4-4x+x^2}\times\left(\frac{2}{2-x}-\frac{8}{8+x^3}\times\frac{4-2x+x^2}{2-x}\right)$$=\frac{2+x}{2-x}\times\frac{4-4x+x^2}{4x^2}\times\left(\frac{2}{2-x}-\frac{8}{\left(2+x\right)\left(4-2x+x^2\right)}\times\frac{4-2x+x^2}{2-x}\right)$$=\frac{2+x}{2-x}\times\frac{\left(2-x\right)^2}{4x^2}\times\left(\frac{2\left(2+x\right)}{\left(2+x\right)\left(2+x\right)}-\frac{8}{\left(2+x\right)\left(2-x\right)}\right)$$=\frac{\left(2+x\right)\left(2-x\right)}{4x^2}\times\frac{4+2x-8}{\left(2+x\right)\left(2-x\right)}$$=\frac{2\left(2+x-4\right)}{4x^2}$$=\frac{x-2}{2x^2}$  Câu trả lời: $\frac{2+x}{2-x}\div\frac{4x^2}{4-4x+x^2}\times\left(\frac{2}{2-x}-\frac{8}{8+x^3}\times\frac{4-2x+x^2}{2-x}\right)$$=\frac{2+x}{2-x}\times\frac{4-4x+x^2}{4x^2}\times\left(\frac{2}{2-x}-\frac{8}{\left(2+x\right)\left(4-2x+x^2\right)}\times\frac{4-2x+x^2}{2-x}\right)$$=\frac{2+x}{2-x}\times\frac{\left(2-x\right)^2}{4x^2}\times\left(\frac{2\left(2+x\right)}{\left(2+x\right)\left(2+x\right)}-\frac{8}{\left(2+x\right)\left(2-x\right)}\right)$$=\frac{\left(2+x\right)\left(2-x\right)}{4x^2}\times\frac{4+2x-8}{\left(2+x\right)\left(2-x\right)}$$=\frac{2\left(2+x-4\right)}{4x^2}$$=\frac{x-2}{2x^2}$

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