Câu hỏi của Bùi Bích Loan

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( 1/x-1 + 1/x+1 ) × ( x - 1/x )

HT.Phong (9A5) · Accepted Answer

$\left(\dfrac{1}{x-1}+\dfrac{1}{x+1}\right)\cdot\left(x-\dfrac{1}{x}\right)$ (1) ĐK: $\left\{{}\begin{matrix}x-1\ne0\x+1\ne0\x\ne0\end{matrix}\right.$ $\Leftrightarrow\left\{{}\begin{matrix}x\ne1\x\ne-1\x\ne0\end{matrix}\right.$ $\Leftrightarrow x\ne\pm1;x\ne0$$\left(1\right)=\left(\dfrac{1}{x-1}+\dfrac{1}{x+1}\right)\cdot\left(\dfrac{x^2}{x}-\dfrac{1}{x}\right)$$=\left(\dfrac{1}{x-1}+\dfrac{1}{x+1}\right)\cdot\dfrac{x^2-1}{x}$$=\left(\dfrac{1}{x-1}+\dfrac{1}{x+1}\right)\cdot\dfrac{\left(x+1\right)\left(x-1\right)}{x}$$=\dfrac{\left(x+1\right)\left(x-1\right)}{x}\cdot\dfrac{1}{x-1}+\dfrac{\left(x+1\right)\left(x-1\right)}{x}\cdot\dfrac{1}{x+1}$$=\dfrac{x+1}{x}+\dfrac{x-1}{x}$$=\dfrac{x+1+x-1}{x}$$=\dfrac{2x}{x}$$=2$Câu trả lời: $\left(\dfrac{1}{x-1}+\dfrac{1}{x+1}\right)\cdot\left(x-\dfrac{1}{x}\right)$ (1) ĐK: $\left\{{}\begin{matrix}x-1\ne0\x+1\ne0\x\ne0\end{matrix}\right.$ $\Leftrightarrow\left\{{}\begin{matrix}x\ne1\x\ne-1\x\ne0\end{matrix}\right.$ $\Leftrightarrow x\ne\pm1;x\ne0$$\left(1\right)=\left(\dfrac{1}{x-1}+\dfrac{1}{x+1}\right)\cdot\left(\dfrac{x^2}{x}-\dfrac{1}{x}\right)$$=\left(\dfrac{1}{x-1}+\dfrac{1}{x+1}\right)\cdot\dfrac{x^2-1}{x}$$=\left(\dfrac{1}{x-1}+\dfrac{1}{x+1}\right)\cdot\dfrac{\left(x+1\right)\left(x-1\right)}{x}$$=\dfrac{\left(x+1\right)\left(x-1\right)}{x}\cdot\dfrac{1}{x-1}+\dfrac{\left(x+1\right)\left(x-1\right)}{x}\cdot\dfrac{1}{x+1}$$=\dfrac{x+1}{x}+\dfrac{x-1}{x}$$=\dfrac{x+1+x-1}{x}$$=\dfrac{2x}{x}$$=2$

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