so sánh P= \(\frac{2\left(x+1\right)}{\sqrt{x}}\)+2 với 5
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6 tháng 7 2016
Rút gọn : \(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\) (ĐKXĐ : \(0\le x\ne1\) )
\(=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\frac{\sqrt{x}-1}{\sqrt{x}}=1-\frac{1}{\sqrt{x}}< 1\)
Vậy B < 1
N
13 tháng 9 2019
\(C=\frac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)^2\)
\(=\sqrt{x}-1\)
Ta co:
\(\sqrt{x}-1+\frac{2}{\sqrt{x}}=\frac{x-\sqrt{x}+2}{\sqrt{x}}=\frac{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{7}{4}}{\sqrt{x}}>0\)
\(\Rightarrow\sqrt{x}-1>-\frac{2}{\sqrt{x}}\)
TM
1 tháng 10 2017
\(A=\left(\frac{1}{\sqrt{x}-1}+\frac{1}{x-\sqrt{x}}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left[\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\frac{\sqrt{x}-1}{\sqrt{x}}=1-\frac{1}{\sqrt{x}}< 1\)
TT
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 : \(P>3\)
\(P-5=\frac{2x+2}{\sqrt{x}}-3=\frac{2x-3\sqrt{x}+2}{\sqrt{x}}=\frac{2\left(x-2\sqrt{x}+1\right)+\sqrt{x}}{\sqrt{x}}=\frac{2\left(\sqrt{x}-1\right)^2}{\sqrt{x}}+1>0\)
\(\Rightarrow P>5\)