Question

cho a khác b khác c khác 0 và\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}\)

Tính \(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)

Answer 1

Theo tính chất của dãy tỉ số bằng nhau

Ta có:\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b+b+c+c+a}{a+b+c}=\dfrac{2a+2b+2c}{a+b+c}=\dfrac{2\times\left(a+b+c\right)}{a+b+c}=2\)

\(\Rightarrow a+b=2\times c\)

\(b+c=2\times a\)

\(c+a=2\times b\)

Ta lại có:\(\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)=\left(\dfrac{b}{b}+\dfrac{a}{b}\right)\left(\dfrac{c}{c}+\dfrac{b}{c}\right)\left(\dfrac{a}{a}+\dfrac{c}{a}\right)\)

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

\(=\dfrac{2\times c}{b}\times\dfrac{2\times a}{c}\times\dfrac{2\times b}{a}\)

\(=2\times2\times2=8\)