Question

Cho 3 số dương a, b, c thỏa mãn: \(a+b+c=\dfrac{1}{2}\) và \(a^2+b^2+c^2+ab+bc+ca=\dfrac{1}{6}\)

Tính giá trị biểu thức \(P=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)

Answer 1

\(\left(a+b+c\right)=\dfrac{1}{2}\Leftrightarrow\left(a+b+c\right)^2=\dfrac{1}{4}\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=\dfrac{1}{4}\)

Ta có: \(ab+bc+ac=\left(a^2+b^2+c^2+2ab+2bc+2ac\right)-\left(a^2+b^2+c^2+ab+bc+ac\right)=\dfrac{1}{4}-\dfrac{1}{6}=\dfrac{1}{12}\)

\(a^2+b^2+c^2=\dfrac{1}{6}-\left(ab+bc+ac\right)=\dfrac{1}{6}-\dfrac{1}{12}=\dfrac{1}{12}\)

Suy ra: \(a^2+b^2+c^2=ab+bc+ac\Leftrightarrow a=b=c\)

\(P=\dfrac{3}{2}\)

p/s làm lih tih k chắc đâu:v