\(\frac{x}{x^2-x+1}=\frac{2}{3}\)
\(\Rightarrow3x=2\left(x^2-x+1\right)\)
\(\Leftrightarrow2x^2-2x+2-3x=0\)
\(\Leftrightarrow2x^2-5x+2=0\)
\(\Leftrightarrow2x^2-4x-x+2=0\)
\(\Leftrightarrow2x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\2x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}}\)
Với x = 2 => Q = 4/21
Với x = 1/2 => Q = 4/21 :))
"Trần Nhật Quỳnh" có cách này ngắn gọn hơn nữa.
Ta có:
\(\frac{x}{x^2-x+1}=\frac{2}{3}\) \(\Rightarrow\frac{x^2-x+1}{x}=\frac{3}{2}\)\(\Rightarrow x-1+\frac{1}{x}=\frac{3}{2}\)
\(\Rightarrow x+\frac{1}{x}=\frac{5}{2}\)
Lại có:
\(Q=\frac{x^2}{x^4+x^2+1}\)
\(\frac{1}{Q}=\frac{x^4+x^2+1}{x^2}\)
\(\frac{1}{Q}=x^2+1+\frac{1}{x^2}\)
\(\frac{1}{Q}=\left(x^2+2x^2.\frac{1}{x^2}+\frac{1}{x^2}\right)-2x^2.\frac{1}{x^2}\)
\(\frac{1}{Q}=\left(x+\frac{1}{x}\right)^2-2\)
Vì \(x+\frac{1}{x}=\frac{5}{2}\)nên
\(\frac{1}{Q}=\left(\frac{5}{2}\right)^2-2\)
\(\frac{1}{Q}=\frac{25}{4}-2\)
\(\frac{1}{Q}=\frac{21}{4}\)
\(\Rightarrow Q=\frac{4}{21}\)
Vậy \(Q=\frac{4}{21}\)
