rút gọn biểu thức : M = \(\frac{x}{\sqrt{x}-1}+\frac{2x-\sqrt{x}}{\sqrt{x}-x}\)
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3 tháng 6 2019
ĐK:x>1
M=\(\frac{x-1}{2x}\) .\(\frac{\left(x-\sqrt{x}\right)\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}\right)\left(\sqrt{x}+1\right)}{x-1}\)
=\(\frac{x-1}{2x}\).\(\frac{x\sqrt{x}-x-x+\sqrt{x}-x\sqrt{x}-x-x-\sqrt{x}}{x-1}\)=\(\frac{x-1}{2x}\).\(\frac{-4x}{x-1}\)=-2
Vậy M=-2
26 tháng 11 2017
\(P=\frac{2x+2}{\sqrt{x}}+\frac{x\sqrt{x}+1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\)
\(P=\frac{2x+2}{\sqrt{x}}+\frac{\sqrt{x^3}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\sqrt{x^3}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(P=\frac{2x+2}{\sqrt{x}}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)\(-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(P=\frac{2x+2}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}\)
\(P=\frac{2x+2-x+\sqrt{x}-1-x+\sqrt{x}-1}{\sqrt{x}}\)
\(P=\frac{2\sqrt{x}}{\sqrt{x}}\)
\(P=2\)
vậy \(P=2\)
5 tháng 4 2017
\(M=\left(\frac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}\right).\frac{x-1}{2x+\sqrt{x}-1}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}}{2\sqrt{x}-1}\)
\(\frac{3x\sqrt{x}+2x}{2x\sqrt{x}+x+\sqrt{x}-1}\)
Đặt \(\sqrt{x}=a\)
Khi đó M=\(\frac{a^2}{a-1}+\frac{2a^2-a}{a-a^2}\)( ĐKXĐ : a \(\ne1,a\ne0\))
=> M = \(\frac{a^2}{a-1}+\frac{a\left(2a-1\right)}{a\left(1-a\right)}\)= \(\frac{a^2}{a-1}+\frac{2a-1}{1-a}=\frac{a^2}{a-1}+\frac{1-2a}{a-1}\)
= \(\frac{a^2-2a+1}{a-1}\)=\(\frac{\left(a-1\right)^2}{a-1}=a-1\)= \(\sqrt{x}-1\)
Vậy M = \(\sqrt{x}-1\)
M = \(\frac{x\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\) + \(\frac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
M = \(2\sqrt{x}-1\)