Lời giải:
ĐKXĐ: $x\geq 1; y\geq 1$
Áp dụng BĐT Cô-si:
$x\sqrt{y-1}=\sqrt{x^2(y-1)}=\sqrt{x(xy-x)}\leq \frac{x+(xy-x)}{2}=\frac{xy}{2}(1)$
$2y\sqrt{x-1}=2\sqrt{y^2(x-1)}=2\sqrt{y(xy-y)}\leq y+(xy-y)=xy(2)$
Từ $(1); (2)\Rightarrow x\sqrt{y-1}+2y\sqrt{x-1}\leq \frac{xy}{2}+xy=\frac{3}{2}xy$
Dấu "=" xảy ra khi $x=xy-x$ và $y=xy-y$
$\Leftrightarrow 2x=xy=2y$
$\Leftrightarrow x=y=2$
ĐKXĐ : \(x\ge1;y\ge1\)
Ta có \(x\sqrt{y-1}+2y\sqrt{x-1}=\dfrac{3xy}{2}\)
\(\Leftrightarrow2x\sqrt{y-1}+4y\sqrt{x-1}=3xy\)
\(\Leftrightarrow\left(2x\sqrt{y-1}-xy\right)+\left(4y\sqrt{x-1}-2xy\right)=0\)
\(\Leftrightarrow x\left(2\sqrt{y-1}-y\right)+2y\left(2\sqrt{x-1}-x\right)=0\)
\(\Leftrightarrow\dfrac{x}{2\sqrt{y-1}+y}\left(4y-4-y^2\right)+\dfrac{2y}{2\sqrt{x-1}+x}\left(4x-4-x^2\right)=0\)
\(\Leftrightarrow\dfrac{x}{2\sqrt{y-1}+y}\left(y-2\right)^2+\dfrac{2y}{2\sqrt{x-1}+x}\left(x-2\right)^2=0\) (1)
Dễ thấy \(\dfrac{x}{2\sqrt{y-1}+y}>0;\dfrac{y}{2\sqrt{x-1}+x}>0\forall x;y\ge1\)
nên (1) \(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\y-2=0\end{matrix}\right.\Leftrightarrow x=y=2\)
Vậy x = y = 2 là nghiệm phương trình
