Question

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

Tính giá trị của biểu thức 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:

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

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

\(P=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)+\left(1+\dfrac{c}{a}\right)=2.2.2=8\)

Answer 2

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

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

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

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

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

P=\(\dfrac{b+a}{b}\).\(\dfrac{c+b}{c}\).\(\dfrac{a+c}{a}\)=\(\dfrac{2c}{b}\).\(\dfrac{2a}{c}.\dfrac{2b}{a}\)=8