Question

Chứng minh rằng nếu a;b;c;d là các số lẻ thì thì:

ƯCLN (\(\dfrac{a+b}{2};\dfrac{b+c}{2};\dfrac{c+a}{2}\)) = ƯCLN (a;b;c)

Answer 1

Gọi d là ƯCLN(\(\dfrac{a+b}{2};\dfrac{b+c}{2};\dfrac{c+a}{2}\))(\(d\ne0,d⋮2\))

Ta có:\(\dfrac{a+b}{2}⋮d,\dfrac{b+c}{2}⋮d,\dfrac{c+a}{2}⋮d\)

\(\Rightarrow\dfrac{a+b}{2}+\dfrac{b+c}{2}+\dfrac{c+a}{2}⋮d\)

\(\Rightarrow\dfrac{a+b+b+c+c+a}{2}⋮d\)

\(\Rightarrow a+b+c⋮d\)

\(\Rightarrow a,b,c⋮d\)

\(\Rightarrow\)ƯCLN(a,b,c)=ƯCLN(\(\dfrac{a+b}{2};\dfrac{b+c}{2};\dfrac{c+a}{2}\))

P/S không chắc đâu nhất là 2 bước cuối