Question

Giúp mik vs

Answer 1

\(\dfrac{x}{x^2-x+1}=\dfrac{2}{3}\\ \Rightarrow2x^2-2x+2=3x\\ \Rightarrow2x^2-5x+2=0\\ \Rightarrow\left(2x^2-4x\right)-\left(x-2\right)=0\\ \Rightarrow2x\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)

Thay x=2 vào Q ta có:

\(Q=\dfrac{x^2}{x^4+x^2+1}=\dfrac{2^2}{2^4+2^2+1}=\dfrac{4}{21}\)

Thay x=\(\dfrac{1}{2}\) vào Q ta có:

\(Q=\dfrac{x^2}{x^4+x^2+1}=\dfrac{\left(\dfrac{1}{2}\right)^2}{\left(\dfrac{1}{2}\right)^4+\left(\dfrac{1}{2}\right)^2+1}=\dfrac{4}{21}\)