Question

tìm \(x\in A\) để biểu thức A∈Z

A=\(\dfrac{x^2+3x+5}{x-1}\)

Answer 1

\(A=\dfrac{x^2+3x+5}{x-1}=\dfrac{x^2-x+4x-4+9}{x-1}=\dfrac{x\left(x-1\right)+4\left(x-1\right)+9}{x-1}=\dfrac{\left(x-1\right)\left(x+4\right)}{x-1}+\dfrac{9}{x-1}=x+4+\dfrac{9}{x-1}\)

Để A∈Z thì: \(\dfrac{9}{x-1}\in Z\)

\(\Leftrightarrow x-1\inƯ\left(9\right)\)

\(\Leftrightarrow x-1=\left\{-9;-3;-1;1;3;9\right\}\)

\(\Leftrightarrow x=\left\{-8;-2;0;2;4;10\right\}\)

vậy........