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10 tháng 8 2020

a) \(ĐKXĐ:\) \(x\ne1,x>0\)

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

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

\(=1:\left[\frac{x+2+x-1-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]\)

\(=1:\frac{\sqrt{x}.\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)

Vậy \(P=\frac{x+\sqrt{x}+1}{\sqrt{x}}\left(x\ne1,x>0\right)\)

b) Xét hiệu \(P-3=\frac{x+\sqrt{x}+1}{\sqrt{x}}-3\)

\(=\frac{x+\sqrt{x}+1-3\sqrt{x}}{\sqrt{x}}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\) \(\forall x>0,x\ne1\)

Do đó : \(P>3\)

1 tháng 10 2017

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

\(=\left[\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)

\(=\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\frac{\sqrt{x}-1}{\sqrt{x}}=1-\frac{1}{\sqrt{x}}< 1\)

23 tháng 10 2017

\(\left[\frac{2}{3\sqrt{x}}-\frac{2}{\sqrt{x}+1}.\left(\frac{\sqrt{x}+1}{3\sqrt{x}}-\sqrt{x}-1\right)\right]:\frac{\sqrt{x}-1}{\sqrt{x}}\)

\(=\left[\frac{2}{3\sqrt{x}}-\frac{2}{\sqrt{x}+1}.\left(\frac{\sqrt{x}+1-3x-3\sqrt{x}}{3\sqrt{x}}\right)\right].\frac{\sqrt{x}}{\sqrt{x}-1}\)

\(=\left[\frac{2}{3\sqrt{x}}-\frac{2}{\sqrt{x}+1}.\frac{-3x-2\sqrt{x}+1}{3\sqrt{x}}\right].\frac{\sqrt{x}}{\sqrt{x}-1}\)

\(=\left[\frac{2}{3\sqrt{x}}-\frac{2}{\sqrt{x}+1}.\frac{\left(\sqrt{x}+1\right)\left(-3\sqrt{x}+1\right)}{3\sqrt{x}}\right].\frac{\sqrt{x}}{\sqrt{x}-1}\)

\(=\left[\frac{2}{3\sqrt{x}}-\frac{-6\sqrt{x}+2}{3\sqrt{x}}\right].\frac{\sqrt{x}}{\sqrt{x}-1}\)

\(=\frac{2\sqrt{x}}{\sqrt{x}-1}\)