Bài 1 : ĐK : \(x>3\) ; \(y>5\) ; \(z>4\)
\(\sqrt{x-3}+\sqrt{y-5}+\sqrt{z-4}=20-\dfrac{4}{\sqrt{x-3}}-\dfrac{9}{\sqrt{y-5}}-\dfrac{25}{\sqrt{z-4}}\)
\(\Leftrightarrow\left(\sqrt{x-3}+\dfrac{4}{\sqrt{x-3}}\right)+\left(\sqrt{y-5}+\dfrac{9}{\sqrt{y-5}}\right)+\left(\sqrt{z-4}+\dfrac{25}{\sqrt{z-4}}\right)=20\)
Theo BĐT Cô - Si cho hai số không âm ta có :
\(\left\{{}\begin{matrix}\sqrt{x-3}+\dfrac{4}{\sqrt{x-3}}\ge2\sqrt{\dfrac{4\sqrt{x-3}}{\sqrt{x-3}}}=2\sqrt{4}=4\\\sqrt{y-5}+\dfrac{9}{\sqrt{y-5}}\ge2\sqrt{\dfrac{9\sqrt{y-5}}{\sqrt{y-5}}}=2\sqrt{9}=6\\\sqrt{z-4}+\dfrac{25}{\sqrt{z-4}}\ge2\sqrt{\dfrac{25\sqrt{z-4}}{\sqrt{z-4}}}=2\sqrt{25}=10\end{matrix}\right.\)
\(\Rightarrow\left(\sqrt{x-3}+\dfrac{4}{\sqrt{x-3}}\right)+\left(\sqrt{y-5}+\dfrac{9}{\sqrt{y-5}}\right)+\left(\sqrt{z-4}+\dfrac{25}{\sqrt{z-4}}\right)\ge20\)
\(\Rightarrow\left(\sqrt{x-3}+\dfrac{4}{\sqrt{x-3}}\right)+\left(\sqrt{y-5}+\dfrac{9}{\sqrt{y-5}}\right)+\left(\sqrt{z-4}+\dfrac{25}{\sqrt{z-4}}\right)=20\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-3}=\dfrac{4}{\sqrt{x-3}}\\\sqrt{y-5}=\dfrac{9}{\sqrt{y-5}}\\\sqrt{z-4}=\dfrac{25}{\sqrt{z-4}}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=4\\y-5=9\\z-4=25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=14\\z=29\end{matrix}\right.\left(TM\right)\)
Vậy \(x=7\) ; \(y=14\) ; \(z=29\)
