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Oanh Kim
Nguyễn Lê Phước Thịnh
30 tháng 7 2023 lúc 12:56

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

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

\(=\dfrac{2x\sqrt{2}+2\sqrt{2x}}{2x-1}\cdot\dfrac{2x-1}{2x+1+x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1-2x-\sqrt{2x}-x\sqrt{2}-\sqrt{x}}\)

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