Đáp án:
\(\left[ \begin{array}{l}
x = \left( {\sqrt {\dfrac{{2521}}{{28}}} - \dfrac{{41\sqrt 7 }}{{14}}} \right):\sqrt 7 \\
x = \left( { - \sqrt {\dfrac{{2521}}{{28}}} - \dfrac{{41\sqrt 7 }}{{14}}} \right):\sqrt 7
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
A = 3\left( {2x - 3} \right)\left( {3x + 2} \right) - 4\left( {x + 4} \right)\left( {4x - 3} \right) + 9x\left( {4 - x} \right)\\
= \left( {6x - 9} \right)\left( {3x + 2} \right) - \left( {4x + 16} \right)\left( {4x - 3} \right) + 26x - 9{x^2}\\
= 18{x^2} + 12x - 27x - 18 - 16{x^2} + 12x - 64x + 48 + 26x - 9{x^2}\\
= - 7{x^2} - 41x + 30\\
A = 0\\
\to - 7{x^2} - 41x + 30 = 0\\
\to - \left( {7{x^2} + 41x - 30} \right) = 0\\
\to - \left[ {{{\left( {x\sqrt 7 } \right)}^2} + 2.x\sqrt 7 .\dfrac{{41\sqrt 7 }}{{14}} + {{\left( {\dfrac{{41\sqrt 7 }}{{14}}} \right)}^2} - \dfrac{{2521}}{{28}}} \right] = 0\\
\to - {\left( {x\sqrt 7 + \dfrac{{41\sqrt 7 }}{{14}}} \right)^2} + \dfrac{{2521}}{{28}} = 0\\
\to {\left( {x\sqrt 7 + \dfrac{{41\sqrt 7 }}{{14}}} \right)^2} = \dfrac{{2521}}{{28}}\\
\to \left[ \begin{array}{l}
x\sqrt 7 + \dfrac{{41\sqrt 7 }}{{14}} = \sqrt {\dfrac{{2521}}{{28}}} \\
x\sqrt 7 + \dfrac{{41\sqrt 7 }}{{14}} = - \sqrt {\dfrac{{2521}}{{28}}}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \left( {\sqrt {\dfrac{{2521}}{{28}}} - \dfrac{{41\sqrt 7 }}{{14}}} \right):\sqrt 7 \\
x = \left( { - \sqrt {\dfrac{{2521}}{{28}}} - \dfrac{{41\sqrt 7 }}{{14}}} \right):\sqrt 7
\end{array} \right.
\end{array}\)
