Question

cho p/s:

A=\(\frac{x-5}{x^2+2}\)

a/ Tính A tại x =3

b/ Tìm x thuộc Z để A thuộc Z

Answer 1

\(A=\frac{x-5}{x^2+2}\\ \)

x=3 => \(A=\frac{3-5}{9+2}\\ =>A=\frac{-2}{11}\)

b) A thuộc Z khi \(x-5⋮x^2+2\\ =>\left(x-5\right)\left(x+5\right)⋮x^2+2\\ =>x^2-10⋮x^2+2\\ =>x^2+2-12⋮x^2+2\)

                             =>12chia hết cho x²+2

                             => x²+2 thuộc U(12)

 

Answer 2

a)Tại x=3 \(A=\frac{3-5}{3^2+2}=\frac{-2}{9+2}=\frac{-2}{11}\)

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

\(=1+\frac{x^2+2}{x^2+2}+\frac{1}{x^2+2}=1+1+\frac{1}{x^2+2}=2+\frac{1}{x^2+2}\in Z\)

\(\Rightarrow1⋮x^2+2\)

\(\Rightarrow x^2+2\inƯ\left(1\right)=\left\{1;-1\right\}\)

\(\Rightarrow x^2\in\left\{-1;-3\right\}\)

\(\Rightarrow x\in\left\{\varnothing\right\}\)