Giải thích các bước giải:
\(\begin{array}{l}
a,\\
3x\left( {x - 2020} \right) - x + 2020 = 0\\
\Leftrightarrow 3x\left( {x - 2020} \right) - \left( {x - 2020} \right) = 0\\
\Leftrightarrow \left( {x - 2020} \right)\left( {3x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2020 = 0\\
3x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2020\\
x = \dfrac{1}{3}
\end{array} \right.\\
b,\\
\left( {2x - 3} \right)\left( {3x + 2} \right) - 6{x^2} = 1\\
\Leftrightarrow \left( {6{x^2} + 4x - 9x - 6} \right) - 6{x^2} = 1\\
\Leftrightarrow \left( {6{x^2} - 5x - 6} \right) - 6{x^2} = 1\\
\Leftrightarrow - 5x - 6 = 1\\
\Leftrightarrow - 5x = 7\\
\Leftrightarrow x = - \dfrac{7}{5}\\
c,\\
{\left( {3x - 1} \right)^2} - {\left( {x + 5} \right)^2} = 0\\
\Leftrightarrow \left[ {\left( {3x - 1} \right) - \left( {x + 5} \right)} \right].\left[ {\left( {3x - 1} \right) + \left( {x + 5} \right)} \right] = 0\\
\Leftrightarrow \left( {2x - 6} \right).\left( {4x + 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2x - 6 = 0\\
4x + 4 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = - 1
\end{array} \right.\\
d,\\
4{x^2} - 4x - 35 = 0\\
\Leftrightarrow \left( {4{x^2} - 14x} \right) + \left( {10x - 35} \right) = 0\\
\Leftrightarrow 2x.\left( {2x - 7} \right) + 5.\left( {2x - 7} \right) = 0\\
\Leftrightarrow \left( {2x - 7} \right)\left( {2x + 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2x - 7 = 0\\
2x + 5 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{7}{2}\\
x = - \dfrac{5}{2}
\end{array} \right.
\end{array}\)
Em xem lại bài 1 nhé, hình như e chép đề sai rồi.
