Giải hệ pt căn(5x^2+2xy+2y^2)+...=3x+3y và căn(x+2y+1)+...=2xy+x+5

Xét VT của (1):

\(3VT\)

\(=\sqrt{5x^2+2xy+2y^2}.\sqrt{2^2+2^2+1^2}+\sqrt{2x^2+2xy+5y^2}.\sqrt{2^2+2^2+1^2}\)

\(=\sqrt{\left(x+y\right)^2+4x^2+y^2}.\sqrt{2^2+2^2+1^2}+\sqrt{\left(x+y\right)^2+x^2+4y^2}.\sqrt{2^2+2^2+1^2}\)

\(\ge\left[2\left(x+y\right)+4x+y\right]+\left[2\left(x+y\right)+x+4y\right]=9x+9y\)

\(\Rightarrow VT\ge3x+3y=VT\)

Đẳng thức xảy ra \(\Leftrightarrow...\Leftrightarrow x=y\)

Sau đó thay \(y=x\) vào pt (2) ta được:

\(\sqrt{3x+1}+2\sqrt[3]{19x+8}=2x^2+x+5\)

\(\Leftrightarrow\left(2x^2-\sqrt{3x+1}\right)+\left(x-5-2\sqrt[3]{19x+8}\right)=0\)

\(\Leftrightarrow\dfrac{4x^2-3x-1}{2x^2+\sqrt{3x+1}}+\dfrac{\left(x+5\right)^3-8\left(19x+8\right)}{\left(x-5\right)^2+2\left(x-5\right)\sqrt[3]{19x+8}+4\sqrt[3]{\left(19x+8\right)^2}}=0\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(4x+1\right)}{2x^2+\sqrt{3x+1}}+\dfrac{ \left(x-1\right)\left(x^2+16x-61\right)}{\left(x-5\right)^2+2\left(x-5\right)\sqrt[3]{19x+8}+4\sqrt[3]{\left(19x+8\right)^2}}=0\)

\(\Leftrightarrow\left(x-1\right)\left[\dfrac{4x+1}{2x^2+\sqrt{3x+1}}+\dfrac{x^2+16x-61}{\left(x-5\right)^2+2\left(x-5\right)\sqrt[3]{19x+8}+4\sqrt[3]{\left(19x+8\right)^2}}\right]=0\)

\(\Leftrightarrow x=1\Rightarrow y=1\)