\(\dfrac{9}{\sqrt{x-19}}+\dfrac{16}{\sqrt{y-5}}+\dfrac{25}{\sqrt{z-91}}=24-\sqrt{x-19}-\sqrt{y-5}-\sqrt{z-91}\\ \Leftrightarrow\left(\dfrac{9}{\sqrt{x-19}}+\sqrt{x-19}\right)+\left(\dfrac{16}{\sqrt{y-5}}+\sqrt{y-5}\right)+\left(\dfrac{25}{\sqrt{z-91}}+\sqrt{z-91}\right)=24\)
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 \(\left\{x;y;z\right\}=\left\{28;21;116\right\}\)
