Điều kiện: $
\left\{\begin{matrix}
4x^2 + 5x + 1 \geq 0 \\
x^2-x+1 \geq 0
\end{matrix}\right.
\Leftrightarrow \left[\begin{matrix}
x \geq - \dfrac{1}{4} \\ \\
x \leq -1
\end{matrix}\right.
$
$\sqrt{4x^2+5x+1} = 9x-3 +2 \sqrt{x^2-x+1} \\
\Leftrightarrow \sqrt{4x^2+5x+1} - 2 \sqrt{x^2-x+1} - (9x-3)=0 \\
\Leftrightarrow \dfrac{4x^2 +5x +1 - 4(x^2-x+1)}{\sqrt{4x^2+5x+1} + 2 \sqrt{x^2-x+1}} - (9x-3) = 0 \\
\Leftrightarrow \dfrac{9x-3}{\sqrt{4x^2+5x+1} + 2 \sqrt{x^2-x+1}} - (9x-3) = 0 \\
\Leftrightarrow (9x-3) \left [ \dfrac{1}{\sqrt{4x^2+5x+1} + 2 \sqrt{x^2-x+1}} - 1 \right ] = 0 \\
\Leftrightarrow
\Leftrightarrow \left[\begin{matrix}
x = \dfrac{1}{3} \\
\dfrac{1}{\sqrt{4x^2+5x+1} + 2 \sqrt{x^2-x+1}} - 1 = 0
\end{matrix}\right.
$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
