Rút gọn biểu thức : \(P=\frac{x\sqrt{2}}{2\sqrt{x}+x\sqrt{2}}+\frac{\sqrt{2x}-2}{x-2}\), với x > 0, x \(\ne\) 2
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PT
22 tháng 10 2019
a) P = \(\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}}{\sqrt{x}+2}\right)\)\(.\)\(\frac{x-4}{\sqrt{4x}}\)
= \(\frac{\sqrt{x}.\left(\sqrt{x}+2\right)+\sqrt{x}.\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)\(.\)\(\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\sqrt{4x}}\)
= \(\frac{x+2\sqrt{x}+x-2\sqrt{x}}{\sqrt{4x}}\)
= \(\frac{2x}{2\sqrt{x}}\)= \(\sqrt{x}\)
b) x = \(3-2\sqrt{2}\)=\(2-2\sqrt{2}+1\)= \(\left(\sqrt{2}-1\right)^2\)
Thay x = \(\left(\sqrt{2}-1\right)^2\) vào P ta được
P = \(\sqrt{\left(\sqrt{2}-1\right)^2}\)= \(\sqrt{2}-1\)
31 tháng 5 2020
1) Ta có: \(A=\left(\frac{2}{\sqrt{x}-2}+\frac{3}{2\sqrt{x}+1}-\frac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\frac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
\(=\left(\frac{2}{\sqrt{x}-2}+\frac{3}{2\sqrt{x}+1}-\frac{5\sqrt{x}-7}{2x-4\sqrt{x}+\sqrt{x}-2}\right)\cdot\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\left(\frac{2\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}+\frac{3\left(\sqrt{x}-2\right)}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\frac{5\sqrt{x}-7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right)\cdot\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\frac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\frac{2\sqrt{x}+3}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\frac{5\sqrt{x}}{2\sqrt{x}+1}\)
TN
29 tháng 6 2019
\(A=\frac{\left(x+\sqrt{x^2-2x}\right)^2-\left(x-\sqrt{x^2-2x}\right)^2}{\left(x-\sqrt{x^2-2x}\right)\left(x+\sqrt{x^2-2x}\right)}\)
\(=\frac{2x\times2\sqrt{x^2-2x}}{2x}=2\sqrt{x^2-2x}\)
\(P=\frac{x\sqrt{2}}{\sqrt{2}\sqrt{x}\left(\sqrt{2}+\sqrt{x}\right)}+\frac{\sqrt{2}\left(\sqrt{x}-\sqrt{2}\right)}{\left(\sqrt{x}-\sqrt{2}\right)\left(\sqrt{x}+\sqrt{2}\right)}\)
\(=\frac{\sqrt{x}}{\sqrt{x}+\sqrt{2}}+\frac{\sqrt{2}}{\sqrt{x}+\sqrt{2}}=1\)