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ĐK: \(x\ge0;x\ne1\)
Ta có: \(P=\text{[}\frac{\sqrt{x}-2}{x-1}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\text{]}\left(\frac{1-x}{\sqrt{2}}\right)^2\)
\(=\text{[}\frac{\sqrt{x}-2}{x-1}-\frac{x+\sqrt{x}-2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\text{]}\frac{\left(x-1\right)^2}{2}\)
\(=\left(\sqrt{x}-2-\frac{x+\sqrt{x}-2}{\sqrt{x}+1}\right)\frac{x-1}{2}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(x+\sqrt{x}-2\right)}{\sqrt{x}+1}.\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{2}\)
\(-2\sqrt{x}.\frac{\sqrt{x}-1}{2}\)\(=\sqrt{x}-x\)
bài 2 : ĐKXĐ : \(x\ge0\) và \(x\ne1\)
Rút gọn :\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{5\sqrt{x}-1}{x-1}\)
\(B=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{5\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-5\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{-1}{\sqrt{x}+1}\)
nhầm rồi bạn ơi!
ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne\pm1\end{matrix}\right.\)
Ta có :
\(P=\left(\frac{x+\sqrt{x}+1}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}\right):\frac{1}{x-1}\)
\(=\left(\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}\right):\frac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\left(\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
\(=\frac{x+\sqrt{x}+1+\sqrt{x}+2+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}.\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(=\frac{x+3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}.\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}.\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(=\left(\sqrt{x}+1\right)^2\)
Vậy..