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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 \(x=7\) ; \(y=14\) ; \(z=29\)
b) \(\dfrac{16}{\sqrt{x-3}}+\dfrac{4}{\sqrt{y-1}}+\dfrac{1225}{\sqrt{z-665}}=82-\sqrt{x-3}-\sqrt{y-1}-\sqrt{z-665}\) (*)
Đk: \(\left\{{}\begin{matrix}x>3\\y>1\\z>665\end{matrix}\right.\)
(*) \(\Leftrightarrow\dfrac{16}{\sqrt{x-3}}+\dfrac{4}{\sqrt{y-1}}+\dfrac{1225}{\sqrt{z-665}}=82-\dfrac{x-3}{\sqrt{x-3}}-\dfrac{y-1}{\sqrt{y-1}}-\dfrac{z-665}{\sqrt{z-665}}\)
\(\Leftrightarrow\dfrac{16}{\sqrt{x-3}}+\dfrac{4}{\sqrt{y-1}}+\dfrac{1225}{\sqrt{z-665}}-82+\dfrac{x-3}{\sqrt{x-3}}+\dfrac{y-1}{\sqrt{y-1}}+\dfrac{z-665}{\sqrt{z-665}}=0\)
\(\Leftrightarrow\left(\dfrac{x-3}{\sqrt{x-3}}-\dfrac{8\sqrt{x-3}}{\sqrt{x-3}}+\dfrac{16}{\sqrt{x-3}}\right)+\left(\dfrac{y-1}{\sqrt{y-1}}-\dfrac{4\sqrt{y-1}}{\sqrt{y-1}}+\dfrac{4}{\sqrt{y-1}}\right)+\left(\dfrac{z-665}{\sqrt{z-665}}-\dfrac{70\sqrt{z-665}}{\sqrt{z-665}}+\dfrac{1225}{\sqrt{z-665}}\right)=0\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x-3}-4\right)^2}{\sqrt{x-3}}+\dfrac{\left(\sqrt{y-1}-2\right)^2}{\sqrt{y-1}}+\dfrac{\left(\sqrt{z-665}-35\right)^2}{\sqrt{z-665}}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-3}-4=0\\\sqrt{y-1}-2=0\\\sqrt{z-665}-35=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=19\\y=5\\z=1890\end{matrix}\right.\)
Kl: x=19, y= 5, z=1890
ĐK : \(x\ge3;y\ge1;z\ge665\)
\(\dfrac{16}{\sqrt{x-3}}+\dfrac{4}{\sqrt{y-1}}+\dfrac{1225}{\sqrt{z-665}}=82-\sqrt{x-3}-\sqrt{y-1}-\sqrt{z-665}\)
\(\Leftrightarrow\left(\dfrac{16}{\sqrt{x-3}}+\sqrt{x-3}\right)+\left(\dfrac{4}{\sqrt{y-1}}+\sqrt{y-1}\right)+\left(\dfrac{1225}{\sqrt{z-665}}+\sqrt{z-665}\right)=82\)
Theo BĐT Cô Si cho các số dương ta có :
\(\left\{{}\begin{matrix}\dfrac{16}{\sqrt{x-3}}+\sqrt{x-3}\ge2\sqrt{\dfrac{16\sqrt{x-3}}{\sqrt{x-3}}}=2\sqrt{16}=8\\\dfrac{4}{\sqrt{y-1}}+\sqrt{y-1}\ge2\sqrt{\dfrac{4\sqrt{y-1}}{\sqrt{y-1}}}=2\sqrt{4}=4\\\dfrac{1225}{\sqrt{z-665}}+\sqrt{z-665}\ge2\sqrt{\dfrac{1225\sqrt{z-665}}{\sqrt{z-665}}}=2\sqrt{1225}=70\end{matrix}\right.\)
\(\Rightarrow\left(\dfrac{16}{\sqrt{x-3}}+\sqrt{x-3}\right)+\left(\dfrac{4}{\sqrt{y-1}}+\sqrt{y-1}\right)+\left(\dfrac{1225}{\sqrt{z-665}}+\sqrt{z-665}\right)\ge82\)
Dấu \("="\) hiển nhiên xảy ra khi :
\(\left\{{}\begin{matrix}\dfrac{16}{\sqrt{x-3}}=\sqrt{x-3}\\\dfrac{4}{\sqrt{y-1}}=\sqrt{y-1}\\\dfrac{1225}{\sqrt{z-665}}=\sqrt{z-665}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=16\\y-1=4\\z-665=1225\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=19\\y=5\\z=1890\end{matrix}\right.\)
a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
18 thì có thể làm 20 thì chịu