Ta có : \(x^3+2x^2+3x+2=y^3\)
\(\Leftrightarrow y^3-x^3=2x^2+3x+2\)
Do \(2x^2+3x+2\)
\(=2\left(x^2+\dfrac{3}{2}x+1\right)\)
\(=2\left(x^2+\dfrac{3}{2}x+\dfrac{9}{16}+\dfrac{7}{16}\right)\)
\(=2\left[\left(x+\dfrac{3}{4}\right)^2+\dfrac{7}{16}\right]\)
\(=2\left(x+\dfrac{3}{4}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}>0\forall x\)
\(\Rightarrow y^3-x^3>0\)
\(\Rightarrow y^3>x^3\left(1\right)\)
Lại có : \(\left(x+2\right)^3=x^3+6x^2+12x+8=x^3+2x^2+3x+2+4x^2+9x+6\)
\(=y^3+4x^2+9x+6\)
\(\Rightarrow\left(x+2\right)^3-y^3=4x^2+9x+6\)
Do \(4x^2+9x+6\)
\(=4\left(x^2+\dfrac{9}{4}x+\dfrac{3}{2}\right)\)
\(=4\left(x^2+\dfrac{9}{4}x+\dfrac{81}{72}+\dfrac{3}{8}\right)\)
\(=4\left[\left(x+\dfrac{9}{8}\right)^2+\dfrac{3}{8}\right]\)
\(=4\left(x+\dfrac{9}{8}\right)^2+\dfrac{3}{2}\ge\dfrac{3}{2}>0\forall x\)
\(\Rightarrow\left(x+2\right)^3>y^3\left(2\right)\)
Từ ( 1 ) ; ( 2 )
\(\Rightarrow x^3< y^3< \left(x+2\right)^3\)
\(\Rightarrow y^3=\left(x+1\right)^3\)
\(\Rightarrow\left\{{}\begin{matrix}y=x+1\\y^3=x^3+3x^2+3x+1\end{matrix}\right.\)
Do \(y^3=x^3+2x^2+3x+2\)
\(\Rightarrow x^3+2x^2+3x+2=x^3+3x^2+3x+1\)
\(\Rightarrow1=x^2\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=1+1=2\\y=-1+1=0\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=1,y=2\\x=-1,y=0\end{matrix}\right.\)
