Question

Some body good at toán jup tui

Cho:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2016}\) và a+b+c=2016

Cmr

Trong a;b;c có 1 số = 2016

 

Answer 1

Ta có : \(a+b+c=2016\Rightarrow\frac{1}{a+b+c}=\frac{1}{2016}\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)+\left(\frac{1}{c}-\frac{1}{a+b+c}\right)=0\)

\(\Leftrightarrow\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=0\)

\(\Leftrightarrow\left(a+b\right)\left[\frac{c^2+ac+bc+ab}{abc\left(a+b+c\right)}\right]=0\)

\(\Leftrightarrow\left(a+b\right)\left(c^2+ac+bc+ab\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\) 

\(\Rightarrow\left[\begin{array}{nghiempt}a+b=0\\b+c=0\\c+a=0\end{array}\right.\)

 Nếu a + b = 0 => c = 2016 (1) Nếu b + c = 0 => a = 2016 (2) Nếu a + c = 0 => b = 2016 (3)

Từ (1) , (2) và (3) ta có điều phải chứng minh.