Bài 1 [TEX]\sqrt{x+1}+\sqrt{2-y}=1+\sqrt{2}[/TEX] và [TEX]\sqrt{2-x}+\sqrt{y+1}=1+\sqrt{2}[/TEX]
Bài 2 [TEX]\sqrt{x+y}+\sqrt{x-y}=4[/TEX]
và [TEX]{{x}^{2}}+{{y}^{2}}=128[/TEX]
Bài 1 ĐKXĐ:...
\[\begin{align}
& \sqrt{x+1}+\sqrt{2-y}=1+\sqrt{2}(1) \\
& \sqrt{2-x}+\sqrt{y+1}=1+\sqrt{2}(2) \\
& (1)-(2)\Leftrightarrow \sqrt{x+1}-\sqrt{y+1}+\sqrt{2-y}-\sqrt{2-x}=0 \\
& \Leftrightarrow \frac{x-y}{\sqrt{x+1}+\sqrt{y+1}}+\frac{x-y}{\sqrt{2-y}+\sqrt{2-x}}=0 \\
& \Leftrightarrow (x-y)(\frac{1}{\sqrt{x+1}+\sqrt{y+1}}+\frac{1}{\sqrt{2-y}+\sqrt{2-x}})=0 \\
& \frac{1}{\sqrt{x+1}+\sqrt{y+1}}+\frac{1}{\sqrt{2-y}+\sqrt{2-x}}>0\Leftrightarrow x=y \\
& (1)\Leftrightarrow \sqrt{x+1}+\sqrt{2-x}=1+\sqrt{2} \\
& \sqrt{x+1}=u;\sqrt{2-x}=v \\
& \Rightarrow {{u}^{2}}+{{v}^{2}}=3;u+v=1+\sqrt{2} \\
& \Rightarrow u+v=1+\sqrt{2};uv=\frac{{{(u+v)}^{2}}-({{u}^{2}}+{{v}^{2}})}{2}=\sqrt{2} \\
& u=1;v=\sqrt{2}\text{ or }u=\sqrt{2};u=1 \\
\end{align}\]
Bài 2: ĐKXĐ
\[\begin{align}
& \sqrt{x+y}+\sqrt{x-y}=4 \\
& {{x}^{2}}+{{y}^{2}}=128 \\
& {{(\sqrt{x+y}+\sqrt{x-y})}^{2}}=16 \\
& \Leftrightarrow x+\sqrt{{{x}^{2}}-{{y}^{2}}}=8 \\
& \Leftrightarrow x+\sqrt{2{{x}^{2}}+128}-8=0 \\
& \Leftrightarrow \sqrt{2{{x}^{2}}+128}=8-x \\
& \Leftrightarrow x\le 8;2{{x}^{2}}+128={{x}^{2}}-16x+64 \\
\end{align}\]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
