Question

Cho a,b,c thỏa mãn a + b + c = \(\frac{1}{2}\) và (a + b)(b + c)(c + a) khác 0

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

Answer 1

Ta có: \(a+b+c=\frac{1}{2}\) \(\Rightarrow\left\{\begin{matrix}a=\frac{1}{2}-b-c\\b=\frac{1}{2}-a-c\\c=\frac{1}{2}-a-b\end{matrix}\right.\) hay \(\left\{\begin{matrix}a+b=\frac{1}{2}-c\\b+c=\frac{1}{2}\\a+c=\frac{1}{2}-b\end{matrix}\right.\)

\(P=\frac{\left(2ab+c\right)\left(2bc+a\right)\left(2ac+b\right)}{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}\)

\(=\frac{\left(2ab+\frac{1}{2}-a-b\right)\left(2bc+\frac{1}{2}-b-c\right)\left(2ca+\frac{1}{2}-a-c\right)}{\left(\frac{1}{2}-c\right)\left(\frac{1}{2}-a\right)\left(\frac{1}{2}-c\right)\left(\frac{1}{2}-b\right)\left(\frac{1}{2}-a\right)\left(\frac{1}{2}-b\right)}\)

\(=\frac{2\left(ab+\frac{1}{4}-\frac{1}{2}a-\frac{1}{2}b\right).2\left(bc+\frac{1}{4}-\frac{1}{2}b-\frac{1}{2}c\right).2\left(ca+\frac{1}{4}-\frac{1}{2}a-\frac{1}{2}c\right)}{\left(ac+\frac{1}{4}-\frac{1}{2}a-\frac{1}{2}c\right)\left(bc+\frac{1}{4}-\frac{1}{2}b-\frac{1}{2}c\right)\left(ab+\frac{1}{4}-\frac{1}{2}a-\frac{1}{2}b\right)}\)

\(=2.2.2=8\)

Vậy với \(a+b+c=\frac{1}{2}\) và \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ne0\) thì \(P=8\)

Answer 2

8

Answer 3

8