\(\dfrac{9}{\sqrt{x-19}}+\dfrac{16}{\sqrt{y-5}}+\dfrac{25}{\sqrt{z-91}}=24-\sqrt{x-19}-\sqrt{y-5}-\sqrt{z-91}\)
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3 tháng 7 2021
\(A=\dfrac{1-\sqrt{x}}{\sqrt{x}+2}=\dfrac{3-\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=\dfrac{3}{\sqrt{x}+2}-1\)
Có \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\)\(\Leftrightarrow\dfrac{3}{\sqrt{x}+2}-1\le\dfrac{1}{2}\)\(\Leftrightarrow A\le\dfrac{1}{2}\)
Dấu "=" xảy ra khi x=0 (tm)
Vậy \(A_{max}=\dfrac{1}{2}\)
Bài 2:
Đk: \(x\ge3;y\ge5;z\ge4\)
Pt\(\Leftrightarrow\sqrt{x-3}+\dfrac{4}{\sqrt{x-3}}+\sqrt{y-5}+\dfrac{9}{\sqrt{y-5}}+\sqrt{z-4}+\dfrac{25}{\sqrt{z-4}}=20\)
Áp dụng AM-GM có:
\(\sqrt{x-3}+\dfrac{4}{\sqrt{x-3}}\ge2\sqrt{\sqrt{x-3}.\dfrac{4}{\sqrt{x-3}}}=4\)
\(\sqrt{y-5}+\dfrac{9}{\sqrt{y-5}}\ge6\)
\(\sqrt{z-4}+\dfrac{25}{\sqrt{z-4}}\ge10\)
Cộng vế với vế \(\Rightarrow VT\ge20\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x-3}=\dfrac{4}{\sqrt{x-3}}\\\sqrt{y-5}=\dfrac{9}{\sqrt{y-5}}\\\sqrt{z-4}=\dfrac{25}{\sqrt{z-4}}\end{matrix}\right.\)\(\Leftrightarrow x=7;y=14;z=29\) (tm)
Vậy...
3 tháng 7 2021
I miss you Được em, hoặc trực tiếp nhóm thành HĐT, một vế là tổng các bình phương, vế còn lại bằng 0
15 tháng 8 2018
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 \(x=7\) ; \(y=14\) ; \(z=29\)
BN
27 tháng 8 2017
Ta có :VT-VP=
\(\left(\dfrac{x}{\sqrt{x}+\sqrt{y}}-\dfrac{y}{\sqrt{x}+\sqrt{y}}\right)+\left(\dfrac{y}{\sqrt{y}+\sqrt{z}}-\dfrac{z}{\sqrt{y}+\sqrt{z}}\right)+\left(\dfrac{z}{\sqrt{z}+\sqrt{x}}-\dfrac{x}{\sqrt{z}+\sqrt{x}}\right)\)\(=\dfrac{x-y}{\sqrt{x}+\sqrt{y}}+\dfrac{y-z}{\sqrt{y}-\sqrt{z}}+\dfrac{z-x}{\sqrt{x}+\sqrt{z}}\)
\(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}+\dfrac{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{y}+\sqrt{z}\right)}{\sqrt{y}+\sqrt{z}}+\dfrac{\left(\sqrt{z}-\sqrt{x}\right)\left(\sqrt{z}+\sqrt{x}\right)}{\sqrt{x}+\sqrt{x}}\)\(=\left(\sqrt{x}-\sqrt{y}\right)+\left(\sqrt{y}-\sqrt{z}\right)+\left(\sqrt{z}-\sqrt{x}\right)=0\)
\(\Rightarrow VT=VP\)
Vậy ...
\(\dfrac{9}{\sqrt{x-19}}+\dfrac{16}{\sqrt{y-5}}+\dfrac{25}{\sqrt{z-91}}=24-\sqrt{x-19}-\sqrt{y-5}-\sqrt{z-91}\\ \Leftrightarrow\left(\dfrac{9}{\sqrt{x-19}}+\sqrt{x-19}\right)+\left(\dfrac{16}{\sqrt{y-5}}+\sqrt{y-5}\right)+\left(\dfrac{25}{\sqrt{z-91}}+\sqrt{z-91}\right)=24\)
Áp dụng BDT: Cô-si:
\(\Rightarrow\left(\dfrac{9}{\sqrt{x-19}}+\sqrt{x-19}\right)+\left(\dfrac{16}{\sqrt{y-5}}+\sqrt{y-5}\right)+\left(\dfrac{25}{\sqrt{z-91}}+\sqrt{z-91}\right)\ge2\sqrt{\dfrac{9}{\sqrt{x-19}}\cdot\sqrt{x-19}}+2\sqrt{\dfrac{16}{\sqrt{y-5}}\cdot\sqrt{y-5}}+2\sqrt{\dfrac{25}{\sqrt{z-91}}\cdot\sqrt{z-91}}\\ =2\cdot3+2\cdot4+2\cdot5=24\)Dấu "=" xảy ra khi:\(\left\{{}\begin{matrix}\dfrac{9}{\sqrt{x-19}}=\sqrt{x-19}\\\dfrac{16}{\sqrt{y-5}}=\sqrt{y-5}\\\dfrac{25}{\sqrt{z-91}}=\sqrt{z-91}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-19=9\\y-5=16\\z-91=25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=28\\y=21\\z=116\end{matrix}\right.\)
Vậy các số \(\left\{x;y;z\right\}=\left\{28;21;116\right\}\)