Question

tìm x thuộc Z để \(Q=\frac{x^2-2x+5}{x-2}\)đạt giá trị nguyên

Answer 1

\(Q=\frac{x^2-2x+5}{x-2}=\frac{x\left(x-2\right)+5}{x-2}=\frac{x\left(x-2\right)}{x-2}+\frac{5}{x-2}=x+\frac{5}{x-2}\)

Q đạt giá trị nguyên <=> \(\frac{5}{x-2}\) nguyên <=> \(5⋮\left(x-2\right)\)

<=>\(x-2\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)

<=>\(x\in\left\{-3;1;3;7\right\}\)

Vậy ............

Answer 2

Ta có: \(\frac{x^2-2x+5}{x-2}=\frac{x\left(x-2\right)+5}{x-2}=x+\frac{5}{x-2}\)

Để Q nguyên thì \(5⋮x-2\)\(\Rightarrow x-2\in\left\{-5;-1;1;5\right\}\)\(\Rightarrow x\in\left\{-3;1;3;7\right\}\)

Nhớ bạn!