1,cho $x>0,y>0$ thoả mãn 
$xy+\sqrt{(x^2+1)(y^2+1)}=\sqrt{2008}$
 
$A=x\sqrt{y^2+1}+y\sqrt{x^2+1}$

2, Cho $x>0,y>0,z>0$ thỏa mãn $x\sqrt{x}+y\sqrt{y}+z\sqrt{z}=3\sqrt{xyz}$
Tính $A=(1+\frac{\sqrt{x}}{\sqrt{y}})(1+\frac{\sqrt{y}}{\sqrt{z}})(1+\frac{\sqrt{z}}{\sqrt{x}})$

3, $x>0,y>0$ thỏa mãn $x+y=\frac{5}{2}\sqrt{xy}$
Tính $\frac{x}{y}$ 

4, Nếu $x,y$ thỏa mãn $x\sqrt{1-y^2}+y\sqrt{1-x^2}=1$
Thì $x^2+y^2=1 $

5,Cho $a+b+c=0, abc\neq 0$ Chứng minh rằng: 
 $\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}|$
3. Ta có:
      $x+y=\dfrac{5}{2}\sqrt{xy}$
$\Leftrightarrow 2x-5\sqrt{xy}+2y=0$
$\Leftrightarrow (2\sqrt x-\sqrt y)(\sqrt x-2\sqrt y)=0$
$\Leftrightarrow \left[\begin{array}{l}2\sqrt x=\sqrt y\\\sqrt x=2\sqrt y\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\dfrac{x}{y}=\dfrac{1}{4}\\\dfrac{x}{y}=4\end{array}\right.$
 5)  Bình phương 2 vế ta được:
           $\frac{1}{a^{2}}$+$\frac{1}{b^{2}}$+$\frac{1}{c^{2}}$ = $\frac{1}{a^{2}}$+$\frac{1}{b^{2}}$+$\frac{1}{c^{2}}$+ 2$\left ( \frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc} \right )$
                                     = $\frac{1}{a^{2}}$+$\frac{1}{b^{2}}$+$\frac{1}{c^{2}}$+ 2$\left ( \frac{a+b+c}{abc} \right )$
         Vì a+b+c=0 nên 
                   2$\frac{0}{abc}$=0
            
          Vậy với a+b+c=0  thì $ \sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}$  = $\left| {\frac{1}{a} }+\frac{1}{b}+\frac{1}{c} \right|$

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