x(y+2)=-8.Tìm x thuộc Z

2 câu trả lời

Đáp án: `(x, y) in {(1;-10),(-1;6),(2;-6),(-2; 2),(4;-4),(-4;0),(8;-3),(-8;-1)`

 

Giải thích các bước giải:

Ta có: `x(y+2)=-8`

`=> x` và `(y+2) in Ư (-8)={+-1; +-2; +-4; +-8}`

Ta có bảng sau:

\begin{array}{|c|c|c|}\hline \text{x}&\text{1}&\text{-1}&\text{2}&\text{-2}&\text{4}&\text{-4}&\text{8}&\text{-8}\\\hline \text{y+2}&\text{-8}&\text{8}&\text{-4}&\text{4}&\text{-2}&\text{2}&\text{-1}&\text{1}\\\hline \text{y}&\text{-10}&\text{6}&\text{-6}&\text{2}&\text{-4}&\text{0}&\text{-3}&\text{-1}\\\hline\end{array}

Vậy`,`

`(x, y) in {(1;-10),(-1;6),(2;-6),(-2; 2),(4;-4),(-4;0),(8;-3),(-8;-1)`

 

Ta có : `x.(y+2)=-8`

Vì `x.(y+2)=-8` nên `x` và `y+2` thuộc `Ư(8)={1;-8;8;-1;-2;4;-4;2}`

Ta có bảng sau :

\begin{array}{|c|c|c|}\hline \text{x}&\text{-1}&\text{8}&\text{-8}&\text{}1&\text{-2}&\text{4}&\text{-4}&\text{2}\\\hline \text{y+2}&\text{8}&\text{-1}&\text{1}&\text{-8}&\text{4}&\text{-2}&\text{2}&\text{-4}\\\hline \text{y}&\text{6}&\text{-3}&\text{-1}&\text{-10}&\text{2}&\text{-4}&\text{0}&\text{6}\\\end{array}

Vậy `(x,y)∈{(-1;6);(8;-3);(-8;-1);(1;-10);(-2;2);(4;-4);(-4;0);(2;6)`