Câu hỏi của Nguyễn Ngọc Trâm

Question

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

Khôi Bùi · Accepted Answer

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)$$y^3+2y^2+3y+1=\left(y^3+3y^2+3y+1\right)-y^2=\left(y+1\right)^3-y^2$ Do $y^2\ge0\forall y\Rightarrow\left(y+1\right)^3-y^2\le\left(y+1\right)^3\forall y\left(2\right)$ Từ ( 1 ) ; ( 2 ) $\Rightarrow\left(y-1\right)^3< x^3\le\left(y+1\right)^3$ $\Rightarrow\left[{}\begin{matrix}x^3=\left(y+1\right)^3\left(3\right)\x^3=y^3\left(4\right)\end{matrix}\right.$ Từ ( 3 ) $\Rightarrow x^3=y^3+3y^2+3y+1$ $\Rightarrow y^3+2y^2+3y+1=y^3+3y^2+3y+1$ $\Rightarrow y^2=0$ $\Rightarrow y=0$ $\Rightarrow\left(y+1\right)^3=1$ $\Rightarrow x^3=1$ $\Rightarrow x=1$ Từ ( 4 ) $\Rightarrow y^3+2y^2+3y+1=y^3$ $\Rightarrow2y^2+3y+1=0$ $\Rightarrow2y^2+2y+y+1=0$ $\Rightarrow2y\left(y+1\right)+y+1=0$ $\Rightarrow\left(2y+1\right)\left(y+1\right)=0$ $\Rightarrow\left[{}\begin{matrix}2y+1=0\y+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=-\dfrac{1}{2}\left(L;y\in Z\right)\y=-1\end{matrix}\right.$ $\Rightarrow y^3=-1=x^3$ $\Rightarrow x=-1$ Vậy $\left(x,y\right)\in\left\{\left(-1,-1\right);\left(1,0\right)\right\}$Câu trả lời: Ta có : $x^3-y^3-2y^2-3y-1=0$ $\Leftrightarrow x^3-\left(y^3+2y^2+3y+1\right)=0$ $\Leftrightarrow x^3=y^3+2y^2+3y+1$ Lại có : $y^3+2y^2+3y+1=\left(y^3-3y^2+3y-1\right)+5y^2+2=\left(y-1\right)^3+5y^2+2$ Do $5y^2\ge0\forall y\Rightarrow\left(y-1\right)^3+5y^2+2\ge\left(y-1\right)^3+2>\left(y-1\right)^3\left(1\right)$$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 $\left(x,y\right)\in\left\{\left(-1,-1\right);\left(1,0\right)\right\}$

nguyen thi minh nguyet · Answer

5y^2 +2 ơ đau zCâu trả lời: 5y^2 +2 ơ đau z

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