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15 tháng 7 2016
Ta có : \(\frac{2}{x^2-y^2}.\sqrt{\frac{3\left(x+y\right)^2}{2}}=\frac{2\sqrt{3}.\left|x+y\right|}{\sqrt{2}.\left(x-y\right)\left(x+y\right)}\)
Vì \(x\ge y\ge0\) nên ta có : \(\left|x+y\right|=x+y\)
\(\Rightarrow\frac{2\sqrt{3}\left|x+y\right|}{\sqrt{2}\left(x-y\right)\left(x+y\right)}=\frac{\sqrt{2}.\sqrt{6}\left(x+y\right)}{\sqrt{2}\left(x-y\right)\left(x+y\right)}=\frac{\sqrt{6}}{x-y}\)
4 tháng 6 2017
P\(=\left(\frac{x-\sqrt{x}+1-x}{x-\sqrt{x}+1}\right).\left(\frac{\sqrt{x^3}+1}{x+2\sqrt{x}+1}\right) \)
\(=\frac{1-\sqrt{x}}{x-\sqrt{x}+1}.\frac{\left(\sqrt{x}+1\right).\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2}\)
\(\frac{1-\sqrt{x}}{\sqrt{x}+1}\)
15 tháng 8 2016
a) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
b) \(\frac{x-1}{\sqrt{y}-1}\cdot\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}+1}\cdot\sqrt{\frac{\left(\sqrt{y}-1\right)^4}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}+1}\cdot\frac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)
15 tháng 8 2016
a)\(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\sqrt{\left(\sqrt{x}-1\right)^2}}{\sqrt{\left(\sqrt{x+1}\right)^2}}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)
b)\(\frac{x-1}{\sqrt{y}-1}\sqrt{\frac{\left(y-2\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}\cdot\frac{\sqrt{\left(\sqrt{y}-1\right)^{2^2}}}{\sqrt{\left(x-1\right)^4}}=\frac{x-1}{\sqrt{y}-1}\cdot\frac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}=\frac{\sqrt{y}-1}{x-1}\)
20 tháng 8 2016
a/ Sai đề.
\(x+2\sqrt{2x-4}=\left(x-2\right)+2.\sqrt{2}.\sqrt{x-2}+2=\left(\sqrt{2}+\sqrt{x-2}\right)^2\)
b/ \(M=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}=\sqrt{\left(\sqrt{2}+\sqrt{x-2}\right)^2}+\sqrt{\left(\sqrt{2}-\sqrt{x-2}\right)^2}\)
\(=\sqrt{2}+\sqrt{x-2}+\left|\sqrt{2}-\sqrt{x-2}\right|\)
1. Nếu \(2\le x\le4\) thì \(M=\sqrt{2}+\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}=2\sqrt{2}\)
2. Nếu \(x>4\) thì \(M=\sqrt{2}+\sqrt{x-2}+\sqrt{x-2}-\sqrt{2}=2\sqrt{x-2}\)
NV
14 tháng 7 2016
a/ \(P=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}-\frac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-\left(3-11\sqrt{x}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)
b/ \(P< 1\Rightarrow\frac{3\sqrt{x}}{\sqrt{x}-3}< 1\Rightarrow\frac{2\sqrt{x}+3}{\sqrt{x}-3}< 0\)
Xét 2 trường hợp:
- \(\hept{\begin{cases}2\sqrt{x}+3>0\\\sqrt{x}-3< 0\end{cases}\Rightarrow\hept{\begin{cases}2\sqrt{x}>-3\\\sqrt{x}< 3\end{cases}\Rightarrow}\hept{\begin{cases}\sqrt{x}>-\frac{3}{2}\\\sqrt{x}< 3\end{cases}}\Rightarrow-\frac{3}{2}< \sqrt{x}< 3}\)
\(\Rightarrow-\frac{9}{4}< x< 9\)
- \(\hept{\begin{cases}2\sqrt{x}+3< 0\\\sqrt{x}>3\end{cases}\Rightarrow\hept{\begin{cases}\sqrt{x}< -\frac{3}{2}\\\sqrt{x}>3\end{cases}}}\) (vô lí)
Vậy -9/4 < x < 9
21 tháng 8 2016
a) \(=\frac{x^2-\sqrt{3^2}}{x+\sqrt{3}}=\frac{\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)}{x+\sqrt{3}}=x-\sqrt{3}\)
LJ
21 tháng 8 2016
\(=\frac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}=a+\sqrt{a+1}\)
9 tháng 7 2016
Ta có : \(A=\frac{\sqrt{x-2\sqrt{2x-4}}}{\sqrt{2}}=\frac{\sqrt{2x-4\sqrt{2x-4}}}{2}=\frac{\sqrt{\left(2x-4\right)-4\sqrt{2x-4}+4}}{2}=\frac{\sqrt{\left(\sqrt{2x-4}-2\right)^2}}{2}=\frac{\left|\sqrt{2x-4}-2\right|}{2}\)
- Với \(2\le x< 4\Rightarrow\sqrt{2x-4}-2< 0\Rightarrow A=\frac{\left|\sqrt{2x-4}-2\right|}{2}=\frac{2-\sqrt{2x-4}}{2}\)
- Với \(x\ge4\Rightarrow\sqrt{2x-4}-2\ge0\Rightarrow A=\frac{\left|\sqrt{2x-4}-2\right|}{2}=\frac{\sqrt{2x-4}-2}{2}\)
Với x >= 0
\(3x+2-\sqrt{x^2}=3x+2-\left|x\right|=3x+2-x=2x+2\)