Question

Tìm x, y biết :

                   |x - 5| + |1 - x| = \(\dfrac{12}{\left|y+1\right|+3}\)

Answer 1

Ta có: \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)

Dấu '=' xảy ra <=> \(ab\ge0\)

Lại có: \(\dfrac{12}{\left|y+1\right|+3}\le\dfrac{12}{3}=4\Rightarrow\left|x-5\right|+\left|1-x\right|\ge4\ge\dfrac{12}{\left|y+1\right|+3}\)

Đẳng thức xảy ra <=> \(\left(x-5\right)\left(1-x\right)\ge0;y+1=0\Rightarrow y=-1\) 

\(x\in Z\Rightarrow x\in\left\{5;4;3;2;1\right\}\)

Vậy ta có cặp số nguyên (x;y)=(5;-1),(4;-1),(3;-1),(2;-1),(1;-1)