Question

[Nguyễn Ngọc Trâm]

Giải pt nghiệm nguyên : x³-y³-2y²-3y-1 =0

Answer 1

[Khôi Bùi]

Ta có : \(x^3-y^3-2y^2-3y-1=0\)

\(\Leftrightarrow x^3-\left(y^3+2y^2+3y+1\right)=0\)

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

Lại có :

\(y^3+2y^2+3y+1=\left(y^3-3y^2+3y-1\right)+5y^2+2=\left(y-1\right)^3+5y^2+2\)

Do \(5y^2\ge0\forall y\Rightarrow\left(y-1\right)^3+5y^2+2\ge\left(y-1\right)^3+2>\left(y-1\right)^3\left(1\right)\)\(y^3+2y^2+3y+1=\left(y^3+3y^2+3y+1\right)-y^2=\left(y+1\right)^3-y^2\)

Do \(y^2\ge0\forall y\Rightarrow\left(y+1\right)^3-y^2\le\left(y+1\right)^3\forall y\left(2\right)\)

Từ ( 1 ) ; ( 2 )

\(\Rightarrow\left(y-1\right)^3< x^3\le\left(y+1\right)^3\)

\(\Rightarrow\left[{}\begin{matrix}x^3=\left(y+1\right)^3\left(3\right)\\x^3=y^3\left(4\right)\end{matrix}\right.\)

Từ ( 3 )

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

\(\Rightarrow y^3+2y^2+3y+1=y^3+3y^2+3y+1\)

\(\Rightarrow y^2=0\)

\(\Rightarrow y=0\)

\(\Rightarrow\left(y+1\right)^3=1\)

\(\Rightarrow x^3=1\)

\(\Rightarrow x=1\)

Từ ( 4 )

\(\Rightarrow y^3+2y^2+3y+1=y^3\)

\(\Rightarrow2y^2+3y+1=0\)

\(\Rightarrow2y^2+2y+y+1=0\)

\(\Rightarrow2y\left(y+1\right)+y+1=0\)

\(\Rightarrow\left(2y+1\right)\left(y+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2y+1=0\\y+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=-\dfrac{1}{2}\left(L;y\in Z\right)\\y=-1\end{matrix}\right.\)

\(\Rightarrow y^3=-1=x^3\)

\(\Rightarrow x=-1\)

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

Answer 2

[nguyen thi minh nguyet]

5y^2 +2 ơ đau z