Question

CHỨNG MINH : \(a\varepsilonℤ\)

thì \((\frac{a}{3}+\frac{a}{2}+\frac{a}{6})\varepsilonℤ\)

Answer 1

Ta có \(\frac{a}{3}+\frac{a}{2}+\frac{a}{6}=\frac{2a}{6}+\frac{3a}{6}+\frac{a}{6}=\frac{6a}{6}=a\)

Mà a thuộc z nên\(\frac{a}{3}+\frac{a}{2}+\frac{a}{6}\)thuộc Z

Answer 2

Ta có : a/3+a/2+a/6

=2a/6+3a/6+a/6

=2a+3a+a/6

=6a/6=a thuộc Z

Answer 3

Ta có:

\(\frac{a}{3}+\frac{a}{2}+\frac{a}{6}=\frac{2a}{6}+\frac{3a}{6}+\frac{a}{6}=\frac{2a+3a+a}{6}=\frac{6a}{6}=a\)

VẬY nếu \(a\in Z\)thì \(\left(\frac{a}{3}+\frac{a}{2}+\frac{a}{6}\right)\in Z\) 

VÌ \(\frac{a}{3}+\frac{a}{2}+\frac{a}{6}=a\)