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rút gọn BT:
a/ \(\sqrt{x-2\sqrt{x-1}}\) (x>2)
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KT
28 tháng 7 2018
ĐK: \(x>0;x\ne1\)
\(A=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
\(A>-1\) \(\Rightarrow\)\(\frac{\sqrt{x}-1}{\sqrt{x}}>-1\)
\(\Leftrightarrow\)\(\frac{\sqrt{x}-1}{\sqrt{x}}+1>0\) \(\Leftrightarrow\)\(\frac{2\sqrt{x}-1}{\sqrt{x}}>0\)
Do \(\sqrt{x}>0\) \(\Rightarrow\)\(2\sqrt{x}-1>0\)\(\Leftrightarrow\)\(2\sqrt{x}>1\)\(\Leftrightarrow\)\(\sqrt{x}>\frac{1}{2}\)\(\Leftrightarrow\)\(x>\frac{1}{4}\)
Vậy \(x>\frac{1}{4}\)\(\left(x\ne1\right)\)thì A > - 1
NN
21 tháng 3 2020
\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne1\end{cases}}\)
Ta có: \(A=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)\(=\left[\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right]:\frac{\sqrt{x}+1}{\left(\sqrt{x}\right)^2-2\sqrt{x}+1}\)
\(=\left[\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right]:\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\frac{\sqrt{x}-1}{\sqrt{x}}\)
Để \(A>-1\)thì \(\frac{\sqrt{x}-1}{\sqrt{x}}>-1\)\(\Leftrightarrow\sqrt{x}-1>-\sqrt{x}\)\(\Leftrightarrow2\sqrt{x}>1\)
\(\Leftrightarrow\sqrt{x}>\frac{1}{2}\)\(\Leftrightarrow x>\frac{1}{4}\)thoả mãn \(x\ne1\)
Vậy \(A>-1\)\(\Leftrightarrow x>\frac{1}{4}\)thoả mãn \(x\ne1\)
NC
30 tháng 8 2020
a) Ta có:
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right)\div\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)
\(P=\frac{x-1}{\left(\sqrt{x}-1\right)\sqrt{x}}\div\frac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(P=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\sqrt{x}}\cdot\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(P=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}=\frac{x-1}{\sqrt{x}}\)
NC
30 tháng 8 2020
b) Ta có: \(P>0\)
\(\Leftrightarrow\frac{x-1}{\sqrt{x}}>0\)
\(\Leftrightarrow\frac{\left(x-1\right)\sqrt{x}}{x}>0\)
\(\Rightarrow\left(x-1\right)\sqrt{x}>0\)
\(\Rightarrow\hept{\begin{cases}x-1>0\\\sqrt{x}>0\end{cases}}\Rightarrow x>1\)
Vậy khi \(x>1\Leftrightarrow P>0\)
c) Ta có: \(P=6\)
\(\Leftrightarrow\frac{x-1}{\sqrt{x}}=6\)
\(\Leftrightarrow x-1=6\sqrt{x}\)
\(\Leftrightarrow\left(x-1\right)^2=36x\)
\(\Leftrightarrow x^2-38x+1=0\)
\(\Leftrightarrow\left(x^2-38x+361\right)-360=0\)
\(\Leftrightarrow\left(x-19\right)^2-\left(6\sqrt{10}\right)^2=0\)
\(\Leftrightarrow\left(x-19-6\sqrt{10}\right)\left(x-19+6\sqrt{10}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-19-6\sqrt{10}=0\\x-19+6\sqrt{10}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=19+6\sqrt{10}\\x=19-6\sqrt{10}\end{cases}}\)
14 tháng 7 2017
a, ĐK \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(Q=\left(1+\frac{\sqrt{x}}{x+1}\right):\left(\frac{1}{\sqrt{x}-1}-\frac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{x+\sqrt{x}+1}{x+1}:\frac{x+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+1\right)}\)\(=\frac{x+\sqrt{x}+1}{x+1}.\frac{\left(\sqrt{x}-1\right)\left(x+1\right)}{\left(\sqrt{x}-1\right)^2}=\frac{x+\sqrt{x}+1}{\sqrt{x}-1}\)
b. \(Q>1\Rightarrow Q-1>0\Rightarrow\frac{x+\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}>0\)
\(\Rightarrow\frac{x+2}{\sqrt{x}-1}>0\)
TH1 \(\hept{\begin{cases}x+2>0\\\sqrt{x}-1>0\end{cases}\Rightarrow\hept{\begin{cases}x>-2\\x>1\end{cases}\Rightarrow}x>1}\)
TH2 \(\hept{\begin{cases}x+2< 0\\\sqrt{x}-1< 0\end{cases}\Rightarrow\hept{\begin{cases}x< -2\\0\le x< 1\end{cases}\left(l\right)}}\)
Vậy \(x>1\)thì \(Q>1\)
PN
9 tháng 8 2016
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{\sqrt{x}-1}\)
Biểu thức \(A\) có nghĩa khi \(\hept{\begin{cases}\sqrt{x}+1\ne0;\text{ }x\ge0\\\sqrt{x}-1\ne0\end{cases}}\) \(\Leftrightarrow\) \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
Ta có:
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{\sqrt{x}-1}=\frac{\sqrt{x}\left(\sqrt{x}+1\right)-2\left(\sqrt{x}-1\right)-2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\frac{x+\sqrt{x}-2\sqrt{x}+2-2\sqrt{x}-2}{x-1}=\frac{x-3\sqrt{x}}{x-1}\)
Vậy, \(A=\frac{x-3\sqrt{x}}{x-1}\)
29 tháng 7 2018
A = \(\frac{\left(x-\sqrt{x}\right)+x-1}{x-\sqrt{x}}\) = \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)+ 1 = \(\frac{\sqrt{x}+1}{\sqrt{x}}\)+ 1 = 2 + \(\frac{1}{\sqrt{x}}\)
Do \(\sqrt{x}\)> 0 mà \(\frac{1}{\sqrt{x}}\)là số nguyên => \(\sqrt{x}\)= 1 => x = 1
26 tháng 7 2018
A = \(\frac{1+x}{x+\sqrt{x}}.\frac{\sqrt{x}+1}{3}\)=\(\frac{1+x}{3\sqrt{x}}\)
ĐKXĐ : x > 0
19 tháng 7 2017
câu 2
\(...=\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2-\sqrt{5}\right|-\left|2+\sqrt{5}\right|=-4\)
câu 1
\(P=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3}{\left(3-\sqrt{x}\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
\(P< -1\Leftrightarrow\frac{-3\sqrt{x}}{2\sqrt{x}+4}+1< 0\Leftrightarrow-\sqrt{x}+4< 0\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\)
2 tháng 6 2019
Rút gọn
\(\frac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}y}=\frac{\sqrt{x}^3\sqrt{y}-xy+yx-\sqrt{x}\sqrt{y}^3}{\sqrt{x}y}=\frac{x}{\sqrt{y}}-y\)
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