2 x 6 - 2 + 3 - 1 - 1 - 1 + 5

= 2 x 6 - ( 2 + 3 + 5 ) - ( 1 - 1 - 1 )

= 12 - 10 - 0

= 2

Ta có : \(a=\frac{x}{x^2-x+1}\Rightarrow\frac{1}{a}=\frac{x^2-x+1}{x}\)

\(\Rightarrow\frac{1}{a^2}=\frac{x^4+x^2+1-2x^3+2x^2-2x}{x^2}\)

\(\Rightarrow\frac{1}{a^2}=\frac{x^4+x^2+1}{x^2}-\frac{2x\left(x^2-x+1\right)}{x^2}\)(1)

mà \(A=\frac{x^2}{x^4+x^2+1}\Rightarrow\frac{1}{A}=\frac{x^4+x^2+1}{x^2}\)

\(\left(1\right)\Leftrightarrow\frac{1}{a^2}=\frac{1}{A}-2.\frac{x^2-x+1}{x}\)

        \(\Leftrightarrow\frac{1}{a^2}=\frac{1}{A}-2.\frac{1}{a}\)

        \(\Leftrightarrow\frac{1}{A}=\frac{1}{a^2}+\frac{2}{a}=\frac{2a+1}{a^2}\)

         \(\Rightarrow A=\frac{a^2}{2a+1}\)