Question

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

tính giá trị của biểu thức P=\(\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(a+\dfrac{c}{a}\right)\)

Answer 1

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

ADTCDTSBN ta có:

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

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

Do đó: \(\dfrac{a+b}{c}=2=>a+b=2c\)

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

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

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

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

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

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

P=2.2.2=8