Question

Cho a, b, c đôi một khác nhau thỏa mãn: ab + bc + ca = 1.Tính giá trị của biểu thức:

A= \(\dfrac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}\)

Answer 1

Ta có:

\(ab+bc+ca=1\)

\(\Rightarrow ab+bc+ca+a^2=a^2+1\)

\(\Rightarrow b\left(a+c\right)+a\left(c+a\right)=a^2+1\)

\(\Rightarrow\left(a+b\right)\left(c+a\right)=a^2+1\)

Tương tự \(b^2+1=\left(a+b\right)\left(b+c\right)\)

\(c^2+1=\left(c+a\right)\left(b+c\right)\)

Suy ra: \(A=\dfrac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}\)

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

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

\(A=1\)