Giải thích các bước giải:
ĐKXĐ: x≥0
$\begin{array}{l} \frac{{{x^2} + x + 1}}{{\sqrt {{x^2} - x + 1} }} = 3\sqrt x \\ \Leftrightarrow {x^2} + x + 1 = 3\sqrt x .\sqrt {{x^2} - x + 1} \\ \Leftrightarrow {(\sqrt {{x^2} - x + 1} )^2} + 2{(\sqrt x )^2} = 3\sqrt x .\sqrt {{x^2} - x + 1} \end{array}$
Đặt $\sqrt {{x^2} - x + 1} = a$(a>0) và $\sqrt x = b$(b≥0)
Ta có:
$\begin{array}{l} {a^2} + 2{b^2} - 3ab = 0\\ \Leftrightarrow (a - b)(a - 2b) = 0\\ \Leftrightarrow \left[ \begin{array}{l} a = b\\ a = 2b \end{array} \right. \end{array}$
+) Khi a=b thì:
$\begin{array}{l} \sqrt {{x^2} - x + 1} = \sqrt x \\ \Leftrightarrow \left\{ \begin{array}{l} x \ge 0\\ {x^2} - x + 1 = x \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x \ge 0\\ {x^2} - 2x + 1 = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x \ge 0\\ {(x - 1)^2} = 0 \end{array} \right.\\ \Leftrightarrow x = 1 \end{array}$(tm ĐKXĐ)
+) Khi a=2b thì:
$\begin{array}{l} \sqrt {{x^2} - x + 1} = 2\sqrt x \\ \Leftrightarrow \left\{ \begin{array}{l} x \ge 0\\ {x^2} - x + 1 = 4x \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x \ge 0\\ {x^2} - 5x + 1 = 0 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} x \ge 0\\ {(x - \frac{5}{2})^2} = \frac{{21}}{4} \end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l} x = \frac{{\sqrt {21} + 5}}{2}\\ x = \frac{{ - \sqrt {21} + 5}}{2} \end{array} \right. \end{array}$
