Question

Cho a,b,c đôi một khác nhau (a#b#c) và \(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}\)

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

Answer 1

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b+b+c+c+a}{c+a+b}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)

\(\Rightarrow P=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)=\dfrac{a+b}{b}\cdot\dfrac{b+c}{c}\cdot\dfrac{c+a}{a}=\dfrac{2c}{b}\cdot\dfrac{2a}{c}\cdot\dfrac{2b}{a}=\dfrac{8abc}{abc}=8\)Vậy P = 8

Answer 2

Ta có: \(\frac{a+b}{c}=\frac{b+c}a{}=\frac{a+c}{b}=\frac{a+b+b+c+c+a}{a+b+c}=2 \)

=> a+b=2c

b+c=2a

a+c=2b

=> P=\((1+\frac{a}{b})(1+\frac{b}{c})(1+\frac{c}{a})=\frac{(a+b)(b+c)(c+a) }{bca} =\frac{2a2b2c}{abc} =8\)