Question

Tìm giá trị của x , biết :
\(\sqrt{x^2-x+9}=2x+1\)

Answer 1

Ta có: \(\sqrt{x^2-x+9}=2x+1\)

\(\Leftrightarrow x^2-x+9=\left(2x+1\right)^2\)

\(\Leftrightarrow x^2-x+9=4x^2+4x+1\)

\(\Leftrightarrow x^2-x+9-4x^2-4x-1=0\)

\(\Leftrightarrow-3x^2-5x+8=0\)

\(\Leftrightarrow-3x^2-8x+3x+8=0\)

\(\Leftrightarrow-x\left(3x+8\right)+\left(3x+8\right)=0\)

\(\Leftrightarrow\left(3x+8\right)\left(1-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+8=0\\1-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-8\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-8}{3}\\x=1\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{-8}{3};1\right\}\)