Câu hỏi của Ahihi

Question

Rút gọn phân thứcA=$\left(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}\right):\dfrac{2x-2}{x}$

Ngô Hải Nam · Accepted Answer

$\left(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}\right):\dfrac{2x-2}{x}$$=\left(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x\left(x-3\right)}\right):\dfrac{2x-2}{x}$$=\left(\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}-\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9}{x\left(x-3\right)}\right)\cdot\dfrac{x}{2\left(x-1\right)}$$=\dfrac{x^2-6x+9-x^2+9}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}$$=\dfrac{-6x+18}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\
=\dfrac{-6\left(x-3\right)}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}$$=\dfrac{-6}{x}\cdot\dfrac{x}{2\left(x-1\right)}$$=\dfrac{-3}{x-1}$Câu trả lời: $\left(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x^2-3x}\right):\dfrac{2x-2}{x}$$=\left(\dfrac{x-3}{x}-\dfrac{x}{x-3}+\dfrac{9}{x\left(x-3\right)}\right):\dfrac{2x-2}{x}$$=\left(\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}-\dfrac{x^2}{x\left(x-3\right)}+\dfrac{9}{x\left(x-3\right)}\right)\cdot\dfrac{x}{2\left(x-1\right)}$$=\dfrac{x^2-6x+9-x^2+9}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}$$=\dfrac{-6x+18}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\
=\dfrac{-6\left(x-3\right)}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}$$=\dfrac{-6}{x}\cdot\dfrac{x}{2\left(x-1\right)}$$=\dfrac{-3}{x-1}$

Ngô Hải Nam · Answer

$ĐK:x\ne0,x\ne3$Câu trả lời: $ĐK:x\ne0,x\ne3$

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