Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
3x\left( {12x - 4} \right) - 9x\left( {4x - 3} \right) = 30\\
\Leftrightarrow x\left( {12x - 4} \right) - 3x\left( {4x - 3} \right) = 10\\
\Leftrightarrow 12{x^2} - 4x - 12{x^2} + 9x = 10\\
\Leftrightarrow 5x = 10\\
\Leftrightarrow x = 2\\
2,\\
x\left( {5 - 2x} \right) + 2x\left( {x - 1} \right) = 15\\
\Leftrightarrow 5x - 2{x^2} + 2{x^2} - 2x = 15\\
\Leftrightarrow 3x = 15\\
\Leftrightarrow x = 5\\
3,\\
\left( {12x - 5} \right)\left( {4x - 1} \right) + \left( {3x - 7} \right)\left( {1 - 16x} \right) = 81\\
\Leftrightarrow \left( {48{x^2} - 12x - 20x + 5} \right) + \left( {3x - 48{x^2} - 7 + 112x} \right) = 81\\
\Leftrightarrow 48{x^2} - 32x + 5 - 48{x^2} + 115x - 7 = 81\\
\Leftrightarrow 83x = 83\\
\Leftrightarrow x = 1\\
4,\\
5x\left( {x - 2000} \right) - x + 2000 = 0\\
\Leftrightarrow 5x.\left( {x - 2000} \right) - \left( {x - 2000} \right) = 0\\
\Leftrightarrow \left( {x - 2000} \right).\left( {5x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2000 = 0\\
5x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2000\\
x = \frac{1}{5}
\end{array} \right.\\
5,\\
{x^3} - 13x = 0\\
\Leftrightarrow x.\left( {{x^2} - 13} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} - 13 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} = 13
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \sqrt {13} \\
x = - \sqrt {13}
\end{array} \right.\\
6,\\
2 - 25{x^2} = 0\\
\Leftrightarrow {x^2} = \frac{2}{{25}}\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{\sqrt 2 }}{5}\\
x = - \frac{{\sqrt 2 }}{5}
\end{array} \right.\\
7,\\
{x^2} - x + \frac{1}{4} = 0\\
\Leftrightarrow {x^2} - 2.x.\frac{1}{2} + {\left( {\frac{1}{2}} \right)^2} = 0\\
\Leftrightarrow {\left( {x - \frac{1}{2}} \right)^2} = 0\\
\Leftrightarrow x - \frac{1}{2} = 0\\
\Leftrightarrow x = \frac{1}{2}\\
8,\\
x\left( {x - 2} \right) + x - 2 = 0\\
\Leftrightarrow x\left( {x - 2} \right) + \left( {x - 2} \right) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = - 1
\end{array} \right.
\end{array}\)
