ĐK:$\begin{cases}x\geq 0;y\geq 0\\ 2x-y>0 \end{cases}$(*)+)$y=0(1)\Leftrightarrow \frac{2}{x}+\frac{1}{x}=\frac{2}{\sqrt{2x^{2}}}$(VN)
+)$x=0:(1)$vô nghiệm
+)$x>0;y>0$.Đặt$t=\frac{x}{y}>0$
(1)tt:$\frac{2}{(\sqrt{t}+1)^{2}}+\frac{1}{t+\sqrt{2t-1}}=\frac{2}{1+\sqrt{t(2t-1)}}$(1')
Ta có:$\frac{1}{t+\sqrt{2t-1}}=\frac{2}{2t+2\sqrt{2t-1}}=\frac{2}{(\sqrt{2t-1}+1)^{2}}$
$(1')\Leftrightarrow\frac{1}{(\sqrt{t}+1)^{2}}+\frac{1}{(\sqrt{2t-1}+1)^{2}}=\frac{1}{1+\sqrt{t(2t-1)}}$
Đặt $\begin{cases}a=\sqrt{t} \\ b=\sqrt{2t-1}\end{cases}\Rightarrow \frac{1}{(a+1)^{2}}+\frac{1}{(1+b)^{2}}=\frac{1}{1+ab}$
ÁD BĐT BCS:$(1+ab)(a+b)\geq (\sqrt{a}+\sqrt{ab}.\sqrt{b})^{2}=a.(1+b)^{2}$
$\Rightarrow \frac{1}{(1+b)^{2}}\geq \frac{a}{a+b}.\frac{1}{1+ab}$
TT$\Rightarrow VT \geq VP$
Dấu''='' xra$\Leftrightarrow a=b\Leftrightarrow \sqrt{t}=\sqrt{2t-1}\Leftrightarrow x=y$
Thay vào(2) :$2(x-4)\sqrt{x-3}-(x-6)\sqrt{2x+1}=3(x-2)$
Dùng pp liên hợp$\Rightarrow (x;y)=(4;4) or (x;y)=(12;12)$