Question

Tìm các cặp số nguyên (x;y) thõa mãn

\(\text{|}x-5\text{|}+\text{|}1-x\text{|}=\frac{12}{\text{|}y+1\text{|}+3}\)

Answer 1

Xét \(VT=\left|x-5\right|+\left|1-x\right|\ge\left|x-5+1-x\right|=4\)(1)

Ta có \(\left|y+1\right|\ge0\Leftrightarrow\left|y+1\right|+3\ge3\Rightarrow\frac{12}{\left|y+1\right|+3}\le\frac{12}{3}=4\) nên \(VP\le4\)(2)

Từ (1) ; (2) \(\Rightarrow VP\le4\le VT\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x-5\right)\left(1-x\right)\ge0\\\left|y+1\right|=0\end{cases}\Rightarrow\hept{\begin{cases}1\le x\le5\\y=-1\end{cases}}}\)