Đáp án:
Giải thích các bước giải:
$\begin{array}{l}
\sqrt {3{x^2} + 33} - \left( {2x + 4} \right) + 3\sqrt x - 3 = 0\\
\Leftrightarrow \frac{{3{x^2} + 33 - {{\left( {2x + 4} \right)}^2}}}{{\sqrt {3{x^2} + 33} + 2x + 4}} + \frac{{3\left( {x - 1} \right)}}{{\sqrt x + 1}} = 0\\
\Leftrightarrow \frac{{ - {x^2} - 16x + 17}}{{\sqrt {3{x^2} + 33} + 2x + 4}} + \frac{{3\left( {x - 1} \right)}}{{\sqrt x + 1}} = 0\\
\Leftrightarrow \frac{{3\left( {x - 1} \right)}}{{\sqrt x + 1}} - \frac{{\left( {x - 1} \right)\left( {x + 17} \right)}}{{\sqrt {3{x^2} + 33} + 2x + 4}} = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {\frac{3}{{\sqrt x + 1}} - \frac{{x + 17}}{{\sqrt {3{x^2} + 33} + 2x + 4}}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
\frac{3}{{\sqrt x + 1}} - \frac{{x + 17}}{{\sqrt {3{x^2} + 33} + 2x + 4}} = 0
\end{array} \right.\\
\frac{3}{{\sqrt x + 1}} - \frac{{x + 17}}{{\sqrt {3{x^2} + 33} + 2x + 4}} = 0\\
\Leftrightarrow 3\sqrt {3{x^2} + 33} = \left( {x + 17} \right)\left( {\sqrt x + 1} \right) - \left( {6x + 12} \right)\,\,\left( 1 \right)\\
Ta\,co:3\sqrt {3{x^2} + 33} = 6x - 9\sqrt x + 21\,\,\left( 2 \right)\\
\left( 1 \right)\,va\,\left( 2 \right)\,suy\,ra:\\
6x - 9\sqrt x + 21 = \left( {x + 17} \right)\left( {\sqrt x + 1} \right) - \left( {6x + 12} \right)\\
\Leftrightarrow x\sqrt x - 11x + 26\sqrt x - 16 = 0\\
\Leftrightarrow \left( {\sqrt x - 8} \right)\left( {\sqrt x - 2} \right)\left( {\sqrt x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x = 8\\
\sqrt x = 2\\
\sqrt x = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 64\\
x = 4\\
x = 1
\end{array} \right.\\
Vay\,phuong\,trinh\,co\,nghiem\,x = 1,x = 4,x = 64
\end{array}$
