Question

\(\dfrac{2-\left(ab+bc+ca\right)}{\left(a-\dfrac{4}{3}\right)^2+\left(b-\dfrac{4}{3}\right)^2+\left(c+\dfrac{4}{3}\right)^2}\)Cho a+b+c=2.Tính gtbt:

Answer 1

Lời giải:

Thực hiện khai triển hằng đẳng thức ta có:

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

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

Mà:\(a+b+c=2\Rightarrow (a+b+c)^2=4\)

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

Vậy: \((a-\frac{4}{3})^2+(b-\frac{4}{3})^2+(c-\frac{4}{3})^2=4-2(ab+bc+ac)\)

\(\Rightarrow \frac{2-(ab+bc+ac)}{(a-\frac{4}{3})^2+(b-\frac{4}{3})^2+(c-\frac{4}{3})^2}=\frac{2-(ab+bc+ac)}{4-2(ab+bc+ac)}=\frac{1}{2}\)