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.Đặtt=\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)