Question

Cho a + b + c = 2. Tính giá trị P=\(\frac{2-\left(ab+bc+ca\right)}{\left(a-\frac{4}{3}\right)^2+\left(b-\frac{4}{3}\right)^2+\left(c-\frac{4}{3}\right)^2}\)

Answer 1

Ta có: \(a+b+c=2\)

\(\Rightarrow\left(a+b+c\right)^2=4\)

\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=4\)

\(\Rightarrow ab+bc+ca=\frac{4-\left(a^2+b^2+c^2\right)}{2}\)

\(P=\frac{2-\left(ab+bc+ca\right)}{\left(a-\frac{4}{3}\right)^2+\left(b-\frac{4}{3}\right)^2+\left(c-\frac{4}{3}\right)^2}\)

\(=\frac{2-\frac{4-\left(a^2+b^2+c^2\right)}{2}}{a^2+b^2+c^2-\frac{8}{3}\left(a+b+c\right)+\frac{16}{3}}\)

\(=\frac{\frac{a^2+b^2+c^2}{2}}{a^2+b^2+c^2}=\frac{1}{2}\)