Question

Tìm x, y ∈ Z biết : (2 - x)(x + 1) = |y + 1|

Answer 1

Ta có: \(\left|y+1\right|=\left(2-x\right)\left(x+1\right)>0\Rightarrow2-x;x+1\) cùng dấu

+ TH1: \(2-x< 0;x+1< 0\Rightarrow x>2;x< -1\) ( vô lí)

+ TH2: \(2-x>0;x+1< 0\Rightarrow x< 2;x>-1\Rightarrow x\in\left\{0;1\right\}\)

-) Với x=0 => |y+1|=2 => y=1; y=-3

-) Với x=1 => |y+1|=2 => y=1; y=-3

+ TH3: \(\left(2-x\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\Rightarrow\left|y+1\right|=0\Rightarrow y=-1}\)

Ta có các cặp số nguyên (x;y) thỏa mãn: (0;1) , ( 0;-3) , (1;1) , ( 1;-3) , ( 2;-1) , ( -1;-1)

Answer 2

\(\left|y+1\right|\ge0\) \(\Rightarrow\left(2-x\right)\left(x+1\right)\ge0\Rightarrow\left[{}\begin{matrix}2-x\ge0,x+1\ge0\\2-x\le0,x+1\le0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\-1\ge x\ge2\left(VL\right)\end{matrix}\right.\) \(\Rightarrow-1\le x\le2\) \(\Rightarrow x\in\left\{-1;0;1;2\right\}\)

*Nếu x=-1 \(\Rightarrow\left|y+1\right|=\left(2+1\right)\left(-1+1\right)=0\Rightarrow y+1=0\Rightarrow y=-1\) 

*Nếu x=0 \(\Rightarrow\left|y+1\right|=\left(2-0\right)\left(0+1\right)=2\Rightarrow\left[{}\begin{matrix}y+1=2\\y+1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=1\\y=-3\end{matrix}\right.\)

*Nếu x=1 \(\Rightarrow\left|y+1\right|=\left(2-1\right)\left(1+1\right)=2\Rightarrow\left[{}\begin{matrix}y+1=2\\y+1=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=1\\y=-3\end{matrix}\right.\)

*Nếu x=2\(\Rightarrow\left|y+1\right|=\left(2-2\right)\left(2+1\right)=0\Rightarrow y+1=0\Rightarrow y=-1\)

Vậy...