Question

[Phan Tiến Nghĩa]

a;b;c là các số thực Cho \(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)

CMR:ab²+bc²+ca²=a³+b³+c³

Answer 1

[Trần Thị Mĩ Duyên]

Ta có \(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)

\(\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)

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

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

Tương tự suy ra \(\frac{1}{c}=\frac{1}{b};\frac{1}{b}=\frac{1}{a}\)

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

Ta có \(ab^2+bc^2+ca^2=a^3+b^3+c^3\)(đccm)

Answer 2

[Huyền Trân]

\(\text{Một cách khác}\)

\(\text{Ta có:}\)

\(\frac{ab}{a+b}=\frac{bc}{b+c}\)

\(\Leftrightarrow ab\left(b+c\right)=bc\left(a+b\right)\)

\(\Leftrightarrow ab^2+abc=abc+b^2c\)

\(\Leftrightarrow a=c\left(1\right)\)

\(\frac{bc}{b+c}=\frac{ca}{a+c}\)

\(\Rightarrow bc\left(a+c\right)=ca\left(b+c\right)\)

\(\Rightarrow abc+bc^2=abc+c^2a\)

\(\Rightarrow b=a\left(2\right)\)

\(Từ\)\(\text{(1) và (2)}\)\(\Rightarrow a=b=c\)

\(\text{Ta có :}\)\(ab^2+bc^2+ca^2=a^3+b^3+c^3\)