√3x^2+33 + 3√x = 2x+7 ( ở đây là căn cả cụm 3x^2 + 33 nha

Đáp án:

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