Question

Tìm x \(\in\) Z sao cho

M = \(\frac{x+2}{3}\in Z\)

N = \(\frac{7}{x-1}\in Z\)

P = \(\frac{x+1}{x-1}\in Z\)

Answer 1

\(N=\frac{7}{x-1}\)

=> x-1 thuộc Ư(7)={-1,-7,1,7}

=> n thuộc {0,-6,2,8}

\(P=\frac{x+1}{x-1}\Leftrightarrow P=\frac{x-1+2}{x-1}\Leftrightarrow P=\frac{x-1}{x-1}+\frac{2}{x-1}\Leftrightarrow P=1+\frac{2}{x-1}\)

=> x-1 thuộc Ư(2)={-1,-2,1,2}

=> n thuộc {0,-1,2,3}

Answer 2

\(M=\frac{x+2}{3}\)nguyên

\(\Leftrightarrow x+2⋮3\)

\(\Rightarrow x+2\in B\left(3\right)=\left\{0;\pm3;\pm6;...\right\}\)

\(\Rightarrow x\in\left\{-2;1;-5;4;-8;...\right\}\)

Vậy....