Question

[Phan Thanh Tịnh]

Cho \(a,b,c\in Z;abc\ne0;\frac{a^2+b^2}{2}=ab;\frac{b^2+c^2}{2}=bc;\frac{a^2+c^2}{2}=ac\)

Tính : \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Answer 1

[o0o I am a studious person o0o]

Ta có : \(\frac{a^2+b^2}{2}=ab\Rightarrow a^2+b^2=2ab\)

\(\Rightarrow a^2-ab+b^2=0\Rightarrow\left(a-b\right)^2=0\Rightarrow a=b\)

Tương tự : \(\frac{b^2+c^2}{2}=bc\Rightarrow b=c\)

\(\frac{a^2+c^2}{2}=ac\Rightarrow a=c\)

Áp dụng t/c bắc cầu ta dc : \(a=b=c\)

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3a\times3=9a\)

Answer 2

[✓ ℍɠŞ_ŦƦùM $₦G ✓]

=>a²+b²=2ab

=>a²-2ab+b²=0

=>(a-b)²=0=>a=b

tương tự=>b=c

=>a=b=c

\(\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3a.3=9a\)