Rút gọn bt
\(\frac{\sqrt{x}}{\sqrt{x}-1}\)+ \(\frac{\sqrt{x}}{\sqrt{x-1}}\)-\(\frac{x+1}{x-1}\)
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TB
25 tháng 7 2018
ĐKXĐ: \(x\ge0\)
\(\frac{1}{\sqrt{x}+1}-\frac{3}{x\sqrt{x}+1}+\frac{2}{x-\sqrt{x}+1}\)
\(=\frac{1}{\sqrt{x}+1}-\frac{3}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}+\frac{2}{x-\sqrt{x}+1}\)
\(=\frac{x-\sqrt{x}+1-3+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}}{x-\sqrt{x}+1}\)
TH
13 tháng 10 2015
a/
\(=\left(\frac{1}{\sqrt{x}+3}+\frac{3}{\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}}{\sqrt{x}+3}-\frac{3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\left(\frac{x-3\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}-3}{\sqrt{x}+3}\right)\)
\(=\left(\frac{x-3\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\frac{x-3\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)^2}\)
\(=\frac{x-3\sqrt{x}+3}{x\sqrt{x}-6\text{x}+9\sqrt{x}}\)
\(=\frac{x-3\sqrt{x}+3}{x\sqrt{x}-6\text{x}+9\sqrt{x}}\)
b/ Vậy để P>1 khi BT trên>1
Ta có phương trình tương đương
\(x-3\sqrt{x}+3-x\sqrt{x}+6\text{x}-9>0\)
\(-x\sqrt{x}+7\text{x}-3\sqrt{x}-6>0\)
Giải pt rồi suy ra
tick cho mình nha
23 tháng 7 2018
a)\(P=\left(\frac{1}{\sqrt{x}-1}+\frac{\sqrt{x}}{x-1}\right):\left(\frac{\sqrt{x}}{\sqrt{x}-1}-1\right)ĐK:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}.\)
\(=\left(\frac{\sqrt{x}+1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\left(\frac{\sqrt{x}-1}{\sqrt{x}-\sqrt{x}+1}\right)\)
=\(\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2\sqrt{x}+1}{\sqrt{x}+1}\)
b)P=3/2 <=>\(\frac{2\sqrt{x}+1}{\sqrt{x}+1}=\frac{3}{2}\Leftrightarrow2\sqrt{x}+1=\frac{3}{2}\sqrt{x}+\frac{3}{2}.\)
\(\Leftrightarrow\frac{1}{2}\sqrt{x}=\frac{1}{2}\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)
Với x=1 thoả nãm yêu cầu
NV
23 tháng 7 2016
1/ ĐKXĐ: \(\hept{\begin{cases}x>0\\x\ne4\end{cases}}\)
\(A=\left[\frac{x}{\sqrt{x}\left(x-4\right)}-\frac{6}{3\left(\sqrt{x}-2\right)}+\frac{1}{\sqrt{x}-2}\right]:\left(\frac{x-4+10-x}{\sqrt{x}+2}\right)\)
\(=\left[\frac{\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}-2}\right]:\left(\frac{6}{\sqrt{x}+2}\right)\)
\(=\frac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\frac{\left(\sqrt{x}+2\right)}{6}\)
\(=\frac{-2}{\sqrt{x}-2}.\frac{1}{6}=-\frac{1}{3\left(\sqrt{x}-2\right)}\)
2/ Để \(A>2\Rightarrow\frac{-1}{3\left(\sqrt{x}-2\right)}>2\)\(\Rightarrow6\sqrt{x}-12+1>0\Rightarrow6\sqrt{x}-11>0\Rightarrow\sqrt{x}>\frac{11}{6}\)
\(\Rightarrow x>\frac{121}{36}\)
1 tháng 2 2022
\(P=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}+\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1}{x-1}\)
\(=2+\dfrac{2x+2}{\sqrt{x}}=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
5 tháng 8 2020
ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne0\\\sqrt{x}-1\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ge0\\x\ne0\\x\ne1\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Ta có : \(P=\left(\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+1}{\sqrt{x}-1}\right)\left(\frac{1}{2\sqrt{x}}-\frac{\sqrt{x}}{2}\right)^2\)
=> \(P=\left(\frac{\left(\sqrt{x}-1\right)^2}{x-1}-\frac{\left(\sqrt{x}+1\right)^2}{x-1}\right)\left(\frac{1}{2\sqrt{x}}-\frac{x}{2\sqrt{x}}\right)^2\)
=> \(P=\left(\frac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{x-1}\right)\left(\frac{1-x}{2\sqrt{x}}\right)^2\)
=> \(P=\left(\frac{\left(\sqrt{x}-1+\sqrt{x}+1\right)\left(\sqrt{x}-1-\sqrt{x}-1\right)}{x-1}\right)\left(\frac{1-x}{2\sqrt{x}}\right)^2\)
=> \(P=\left(\frac{-4\sqrt{x}}{x-1}\right)\left(\frac{1-x}{2\sqrt{x}}\right)^2\)
=> \(P=\frac{-4\sqrt{x}\left(x-1\right)^2}{\left(2\sqrt{x}\right)^2\left(x-1\right)}\)
=> \(P=-\frac{x-1}{\sqrt{x}}\)
Vậy ...
Giúp với mình gấp quá
\(=\frac{2\sqrt{x}}{\sqrt{x}-1}-\frac{x+1}{x-1}\)
\(=\frac{2\sqrt{x}\left(x-1\right)-\left(x+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x-1\right)}\)
\(=\frac{2x\sqrt{x}-2\sqrt{x}-x\sqrt{x}+x-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x-1\right)}\)
\(=\frac{x\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x-1\right)}\)
\(=\frac{x\sqrt{x}-\sqrt{x}+1}{x\sqrt{x}-\sqrt{x}-x+1}=1-\frac{x}{\left(\sqrt{x}-1\right)\left(x-1\right)}\)