ĐKXĐ:...
\(\left(x+2\right)^3-y^3+x+2-y=0\)
\(\Leftrightarrow\left(x+2-y\right)\left[\left(x+2\right)^2+y\left(x+2\right)+y^2+1\right]=0\)
\(\Leftrightarrow\left(x+2-y\right)\left[\left(x+2+\frac{y}{2}\right)^2+\frac{3y^2}{4}+1\right]=0\)
\(\Leftrightarrow x+2-y=0\Rightarrow y=x+2\)
Thay vào pt đầu:
\(4\sqrt{x+2}+2\sqrt{3\left(x+4\right)}=3\left(x+2\right)\left(x+1\right)+10\)
\(\Leftrightarrow3x^2+9x+16-4\sqrt{x+2}-2\sqrt{3\left(x+4\right)}=0\)
\(\Leftrightarrow3\left(x^2+2x+1\right)+2\left(x+3-2\sqrt{x+2}\right)+\left(x+7-2\sqrt{3x+12}\right)=0\)
\(\Leftrightarrow3\left(x+1\right)^2+\frac{2\left(x+1\right)^2}{x+3+2\sqrt{x+2}}+\frac{\left(x+1\right)^2}{x+7+2\sqrt{3x+12}}=0\)
\(\Rightarrow x=-1\)
