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\(\left\{{}\begin{matrix}\sqrt{x^2+3}+2\sqrt{x}=3+\sqrt{y}\left(1\right)\\\sqrt{y^2+3}+2\sqrt{y}=3+\sqrt{x}\left(2\right)\end{matrix}\right.\)\(\left(đk;x;y\ge0\right)\)
\(\left(1\right)-\left(2\right)\Rightarrow\sqrt{x^2+3}+2\sqrt{x}-\sqrt{y^2+3}-2\sqrt{y}=\sqrt{y}-\sqrt{x}\)
\(\Leftrightarrow\sqrt{x^2+3}-\sqrt{y^2+3}+2\sqrt{x}-2\sqrt{y}+\sqrt{x}-\sqrt{y}=0\left(3\right)\)
\(với:x=y=0\Rightarrow ko\) \(là\) \(nghiệm\)
\(vỡi:x=y\ne0\Rightarrow x;y>0\)
\(\left(3\right)\Leftrightarrow\dfrac{x^2+3-y^2-3}{\sqrt{x^2+3}+\sqrt{y^2+3}}+\dfrac{4x-4y}{2\sqrt{x}+2\sqrt{y}}+\dfrac{x-y}{\sqrt{x}+\sqrt{y}}=0\)
\(\Leftrightarrow\left(x-y\right)\left[\dfrac{x+y}{\sqrt{x^2+3}+\sqrt{y^2+3}}+\dfrac{4}{2\sqrt{x}+2\sqrt{y}}+\dfrac{1}{\sqrt{x}+\sqrt{y}}>0\left(\forall x;y>0\right)\right]=0\)
\(\Rightarrow x=y\left(4\right)\)
\(\left(4\right)và\left(1\right)\Rightarrow\sqrt{x^2+3}+2\sqrt{x}=3+\sqrt{x}\Leftrightarrow\sqrt{x^2+3}+\sqrt{x}-3=0\)
\(\Leftrightarrow\sqrt{x^2+3}-2+\sqrt{x}-1=0\Leftrightarrow\dfrac{x^2+3-4}{\sqrt{x^2+3}+2}+\dfrac{x-1}{\sqrt{x}+1}=0\Leftrightarrow\left(x-1\right)\left[\dfrac{x+1}{\sqrt{x^2+3}+2}+\dfrac{1}{\sqrt{x}+1}>0\left(\forall x>1\right)\right]=0\Leftrightarrow x=y=1\)
Ta có: \(\left\{{}\begin{matrix}x^3+y^2=2\\x^2+y^3=2\end{matrix}\right.\) \(\Rightarrow x^3+y^2=x^2+y^3\Leftrightarrow x^3-x^2=y^3-y^2\Leftrightarrow x^2\left(x-1\right)=y^2\left(y-1\right)\)
\(\Rightarrow\) x=y
\(\Rightarrow\)\(x^3+y^2=2\Leftrightarrow x^3+x^2=2\Leftrightarrow x=1\)\(\Rightarrow y=1\)
em tưởng trừ 2 vế ra là \(\left(x-1\right)\left(x^2+y^2+xy-x-y\right)=0\)