TH

Rút gọn P=\(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x\sqrt[]{x}+x-\sqrt{x}-1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\)

H24
13 tháng 6 2023 lúc 10:41

\(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x\sqrt{x}+x-\sqrt{x}-1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\left(dkxd:x\ne1,x\ge0\right)\)

\(=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{\sqrt{x}\left(x-1\right)+\left(x-1\right)}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)

\(=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{\sqrt{x}+1-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)

\(=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right).\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)

\(=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\left(\sqrt{x}+1\right)\)

\(=\dfrac{x+2\sqrt{x}+1-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\left(\sqrt{x}+1\right)\)

\(=\dfrac{x+\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x+\sqrt{x}+3}{x-1}\)

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