Đặt \(\left\{{}\begin{matrix}\sqrt{x-2}=a\left(a>0\right)\\\sqrt{y-1}=b\left(b>0\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{36}{a}+\dfrac{4}{b}=28-4a-b\)
\(\Leftrightarrow\left(\dfrac{36}{a}+4a\right)+\left(\dfrac{4}{b}+b\right)=28\)
\(VT\ge2\sqrt{\dfrac{36}{a}\times4a}+2\sqrt{\dfrac{4}{b}\times b}=28\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\dfrac{36}{a}=4a\\\dfrac{4}{b}=b\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\) \(\left(a,b>0\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-2}=3\\\sqrt{y-1}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=11\\y=5\end{matrix}\right.\) (n)
Vậy . . . >3<
