Bạn hỏi ở môn Toán nhé
\(Q=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{\left(1-x\right)^2}{2}.ĐKXĐ:x\ne1\)
\(=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(1-x\right)^2}{2}\)
\(=\left(\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\dfrac{\left(1-x\right)^2}{2}\)
\(=\left(\dfrac{x+\sqrt{x}-2\sqrt{x}-2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}-\dfrac{x-\sqrt{x}+2\sqrt{x}-2}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\right).\dfrac{\left(1-x\right)^2}{2}\)
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