Question

Решите уравнение..............

Answer 1

$\sqrt{x^2-x+1}=y,y\geq 0\Rightarrow (y-2x)^2(y+4x)=0$

$y=2x\Rightarrow \sqrt{x^2-x+1}=2x\Leftrightarrow \begin{cases}x^2-x+1=4x^2\\ x\geq 0\end{cases}\Leftrightarrow \begin{cases}3x^2+x-1=0\\ x\geq 0\end{cases}\\\begin{cases}3\left ( 6x+1 \right )^2=\cfrac{13}{12}\\ x\geq 0\end{cases}\Rightarrow \begin{cases}x=\cfrac{-1\pm \sqrt{13}}{6}\\ x\geq 0\end{cases}\Rightarrow x=\cfrac{-1+\sqrt{13}}{6}$

$y=-4x\Rightarrow \sqrt{x^2-x+1}=-4x\Leftrightarrow \begin{cases}x^2-x+1=16x^2\\ x\leq 0\end{cases}\Leftrightarrow \begin{cases}15x^2+x-1=0\\ x\leq 0\end{cases}\\\begin{cases}15\left ( 30x+1 \right )^2=\cfrac{61}{60}\\ x\leq 0\end{cases}\Rightarrow \begin{cases}x=\cfrac{-1\pm \sqrt{61}}{30}\\ x\leq 0\end{cases}\Rightarrow x=\cfrac{-1-\sqrt{61}}{30}$