Question

Tìm tất cả các cặp số nguyên x,y có tích là 1, sao cho \(\frac{x+y-2}{4}=\frac{1}{x-1}+\frac{1}{y+1}\)

Answer 1

\(\frac{x+y-2}{4}=\frac{y+1+x-1}{\left(x-1\right)\left(y-1\right)}\Leftrightarrow\frac{x+y-2}{4}=\frac{x+y}{xy-x-y+1}\)

\(\Leftrightarrow\frac{x+y-2}{4}=\frac{x+y}{-\left(x+y\right)}=-1\)

\(\Leftrightarrow x+y-2=-4\Rightarrow x+y=-2\Rightarrow y=-2-x\)

Mà \(xy=1\Rightarrow x\left(-2-x\right)=1\Leftrightarrow x^2+2x+1=0\)

\(\Rightarrow\left(x+1\right)^2=0\Rightarrow x=-1\Rightarrow y=-1\)