Question

Giải phương trình nghiệm nguyên

\(x^3+2x^2+3x+2=y^3\)

Answer 1

Ta có:

\(2x^2+3x+2=2\left(x^2+\dfrac{3}{2}x+1\right)=2\left(x+\dfrac{3}{4}\right)^2+\dfrac{7}{8}>0\)

\(\Leftrightarrow2x^2+3x+2>0\)

\(\Leftrightarrow x^3+2x^2+3x+2>x^3\)

\(\Leftrightarrow y^3>x^3\)

\(\Leftrightarrow y>x\) (1)

Chứng minh y<x+2

\(\Leftrightarrow y^3< x^3+6x^2+12x+8\)

\(\Leftrightarrow x^3+2x^2+3x+2< x^3+6x^2+12x+8\)

\(\Leftrightarrow4x^2+9x+6>0\)

\(\Leftrightarrow\left(2x+\dfrac{9}{4}\right)^2+\dfrac{15}{8}>0\left(lđ\right)\)

\(\Leftrightarrow x+2< y\left(2\right)\)

Từ (1) và (2) suy ra

y=x+1

\(\Leftrightarrow y^3=x^3+3x^2+3x+1\)

\(\Leftrightarrow x^3+2x^2+3x+2=x^3+3x^2+3x+1\)

\(\Leftrightarrow x^2-1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

TH1: x=1

\(\Rightarrow y=x+1\Leftrightarrow y=2\)

TH1: x=-1

\(\Rightarrow y=x+1\Leftrightarrow y=0\)

Vậy \(\left(x;y\right)\in\left\{\left(1;2\right);\left(-1;0\right)\right\}\)

Answer 2

x³<y³<(x+1)³=> vô nghiệm