Question

Cho 3 số thực a, b, c khác 0 thõa mãn

\(a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)=-2\)

và \(a^3+b^3+c^3=1\)

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

Answer 1

Ta có:
\(a\left(\frac{1}{b}+\frac{1}{c}\right)+b\left(\frac{1}{c}+\frac{1}{a}\right)+c\left(\frac{1}{a}+\frac{1}{b}\right)=-2\)
\(\Leftrightarrow\frac{a}{b}+\frac{b}{a}+\frac{a}{c}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b}=-2\)
\(\Leftrightarrow\left(\frac{a}{b}+\frac{a}{c}+1\right)+\left(\frac{b}{a}+\frac{b}{c}+1\right)+\left(\frac{c}{a}+\frac{c}{b}+1\right)=1\)
\(\Leftrightarrow a\left(\frac{1}{b}+\frac{1}{c}+\frac{1}{a}\right)+b\left(\frac{1}{a}+\frac{1}{c}+\frac{1}{b}\right)+c\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)(1)
Mặt khác:
\(\left(a+b+c\right)^3=a^3+b^3+c^3+3ab\left(a+b\right)+3bc\left(b+c\right)+3ca\left(c+a\right)+6abc\)
\(\Rightarrow\frac{\left(a+b+c\right)^3}{abc}=\frac{1}{abc}+3\left(\frac{a}{b}+\frac{b}{a}+\frac{a}{c}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b}\right)+6\)
\(\Leftrightarrow\frac{\left(a+b+c\right)^3}{abc}=\frac{1}{abc}+3.\left(-2\right)+6\)
\(\Leftrightarrow\frac{\left(a+b+c\right)^3}{abc}=\frac{1}{abc}\)
\(\Leftrightarrow\left(a+b+c\right)^3=1\)
\(\Leftrightarrow a+b+c=1\)(2)
Từ (1) và (2) \(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\left(đpcm\right)\)

 

Answer 2

bạn cho mình hỏi thê thì dữ liệu a^3+b^3+c^3 không được dùng à

Answer 3

nguyễn thị vân anh có dùng đó bạn, khai triển (a+b+c)³ =a³+b³+c³+... rồi thay a³+b³+c³=1 đó, bạn xem lại bài mik đi