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ĐKXĐ: \(x\ge1\); x khác 2; 3
Ta có:
\(\frac{1}{\sqrt{x}-\sqrt{x-1}}=\frac{\sqrt{x}+\sqrt{x-1}}{x-\left(x-1\right)}=\sqrt{x}+\sqrt{x-1}\)
\(\frac{x-3}{\sqrt{x-1}-\sqrt{2}}=\frac{\left(x-3\right)\left(\sqrt{x-1}+\sqrt{2}\right)}{x-1-2}=\sqrt{x-1}+\sqrt{2}\)
=> \(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{x-3}{\sqrt{x-1}-\sqrt{2}}=\sqrt{x}+\sqrt{x-1}-\left(\sqrt{x-1}+\sqrt{2}\right)=\sqrt{x}-\sqrt{2}\)
\(\frac{2}{\sqrt{2}-\sqrt{x}}-\frac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x}=\frac{2\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}=\frac{\sqrt{x}-2}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}\)
=> \(P=\left(\sqrt{x}-\sqrt{2}\right).\frac{\sqrt{x}-2}{\sqrt{x}\left(\sqrt{2}-\sqrt{x}\right)}=\frac{2-\sqrt{x}}{\sqrt{x}}\)
\(S=\frac{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)+\left(2x-\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-3x\sqrt{x}+2x-\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(S=\frac{x\sqrt{x}-2x+2\sqrt{x}-1+2x\sqrt{x}+x-2\sqrt{x}-1-3x\sqrt{x}+2x-\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(S=\frac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(S=\frac{1}{\sqrt{x}+1}\)
Vậy \(S=\frac{1}{\sqrt{x}+1}\)