Question

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

a, \(a^3+b^3+c^3=3abc\)

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

Answer 1

a,ta có: \(a^3+b^3+c^3=3abc\)

<=>\(a^3+b^3+c^3-3abc=0\)

<=>\(\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)

<=>\(\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)

<=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)

<=>\(\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

<=>\(\left(a+b+c\right)2\left(a^2-ab+b^2-ac-bc+c^2\right)=0\)

<=>\(\left(a+b+c\right)\left(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\right)=0\)

=>a=b,a=c,b=c

=>a=b=c

thay a=b=c vào P ta đc

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