Question

Cho a, b, c là 3 số khác 0 thỏa mãn: \(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)(với giả thiết các tỉ số đều có nghĩa).

Tính giá trị của biểu thức \(N=\dfrac{ab+bc+ca}{a^2+b^2+c^2}\)

Answer 1

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

\(=>\)\(\dfrac{a+b}{ab}\)=\(\dfrac{b+c}{bc}\)=\(\dfrac{c+a}{ca}\)

\(=>\)\(\dfrac{1}{a}\)+\(\dfrac{1}{b}\)=\(\dfrac{1}{b}\)+\(\dfrac{1}{c}\)=\(\dfrac{1}{c}\)+\(\dfrac{1}{a}\)

\(=>\)\(\dfrac{1}{b}\)+\(\dfrac{1}{a}\)=\(\dfrac{1}{c}\)+\(\dfrac{1}{b}\)

\(\dfrac{1}{c}\)+\(\dfrac{1}{b}\)=\(\dfrac{1}{a}\)+\(\dfrac{1}{c}\)

\(\dfrac{1}{a}\)+\(\dfrac{1}{c}\)=\(\dfrac{1}{b}\)+\(\dfrac{1}{a}\)

\(=>\)\(\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}\)

\(=>\)a=b=c

Vậy: M=\(\dfrac{ab+bc+ca}{a^2+b^2+c^2}=\dfrac{a^2+a^2+a^2}{a^2+a^2+a^2}\)

= 1

Answer 2

mình bt nè