Đáp án:
Giải thích các bước giải:
\[\begin{array}{l}
1.\\
1,12x\left( {3x - 1} \right) + \left( {5 - 6x} \right)\left( {5 + 6x} \right)\\
= 36{x^2} - 12x + 25 - 36{x^2} = 25 - 12x\\
2,\left( {2{x^2} - 3} \right).4x - {\left( {3x - 2} \right)^2} = 8{x^3} - 12x - \left( {9{x^2} - 12x + 4} \right)\\
= 8{x^3} - 12x - 9{x^2} + 12x - 4\\
= 8{x^3} - 9{x^2} - 4\\
2.\\
1,\\
{x^2} - 4{y^2} - 10 + 25 = \left( {{x^2} - 10x + 25} \right) - {\left( {2y} \right)^2}\\
= {\left( {x - 5} \right)^2} - {\left( {2y} \right)^2} = \left( {x - 2y - 5} \right)\left( {x + 2y - 5} \right)\\
2,\\
3{x^2} - x - 2 = \left( {3{x^2} - 3x} \right) + \left( {2x - 2} \right)\\
= 3x\left( {x - 1} \right) + 2\left( {x - 1} \right) = \left( {3x + 2} \right)\left( {x - 1} \right)\\
3.\\
1,3{x^3} - 75x = 0\\
\Leftrightarrow {x^3} - 25x = 0\\
\Leftrightarrow x\left( {{x^2} - 25} \right) = 0\\
\Leftrightarrow x\left( {x - 5} \right)\left( {x + 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 5\\
x = - 5
\end{array} \right.\\
2,{\left( {x + 2} \right)^2} - x\left( {x + 3} \right) = 5\\
\Leftrightarrow {x^2} + 4x + 4 - {x^2} - 3x - 5 = 0\\
\Leftrightarrow x - 1 = 0\\
\Leftrightarrow x = 1
\end{array}\]
