Đáp án:
$\begin{array}{l}
1)\\
{x^2} - 7x = 0\\
\Leftrightarrow x\left( {x - 7} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - 7 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 7
\end{array} \right.\\
Vậy\,x = 0;x = 7\\
2){x^3} - 25x = 0\\
\Leftrightarrow x\left( {{x^2} - 25} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} = 25
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 5;x = - 5
\end{array} \right.\\
Vậy\,x = 0;x = - 5;x = 5\\
3){x^4} - 4{x^2} = 0\\
\Leftrightarrow {x^2}\left( {{x^2} - 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} = 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 2\\
x = - 2
\end{array} \right.\\
Vậy\,x = 0;x = 2;x = - 2\\
4)\left( {1 - 2x} \right)\left( {1 + 2x} \right) - x\left( {x + 2} \right)\left( {x - 2} \right) = 0\\
\Leftrightarrow 1 - 4{x^2} - x\left( {{x^2} - 4} \right) = 0\\
\Leftrightarrow 1 - 4{x^2} - {x^3} + 4x = 0\\
\Leftrightarrow {x^3} + 4{x^2} - 4x - 1 = 0\\
\Leftrightarrow {x^3} - {x^2} + 5{x^2} - 5x + x - 1 = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {{x^2} + 5x + 1} \right) = 0\\
\Leftrightarrow x = 1;x = \frac{{ - 5 \pm \sqrt {21} }}{2}\\
Vậy\,x = 1;x = \frac{{ - 5 \pm \sqrt {21} }}{2}\\
5)\left( {x + 5} \right)\left( {4 - 3x} \right) - {\left( {3x + 2} \right)^2} + {\left( {2x + 1} \right)^3}\\
= \left( {2x - 1} \right)\left( {4{x^2} + 2x + 1} \right)\\
\Leftrightarrow 4x - 3{x^2} + 20 - 15x - 9{x^2} - 12x - 4\\
+ 8{x^3} + 6{x^2} + 12x + 1 = 8{x^3} - 1\\
\Leftrightarrow - 6{x^2} - 11x + 18 = 0\\
\Leftrightarrow x = \frac{{ - 11 \pm \sqrt {553} }}{{12}}\\
Vậy\,x = \frac{{ - 11 \pm \sqrt {553} }}{{12}}\\
6){\left( {x - 3} \right)^2} - 5\left( {3 - x} \right) = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {x - 3 + 5} \right) = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow x = 3;x = - 2\\
Vậy\,x = 3;x = - 2\\
7){\left( {x + 3} \right)^2} - \left( {x - 2} \right)\left( {x + 2} \right) - 25 = 0\\
\Leftrightarrow {x^2} + 6x + 9 - {x^2} + 4 - 25 = 0\\
\Leftrightarrow 6x = 12\\
\Leftrightarrow x = 2\\
Vậy\,x = 2\\
8){\left( {2x - 2} \right)^2} - \left( {4x - 1} \right)\left( {x + 3} \right) + 9x = 17\\
\Leftrightarrow 4{x^2} - 8x + 4 - 4{x^2} - 12x + x + 3 + 9x = 17\\
\Leftrightarrow - 10x = 10\\
\Leftrightarrow x = - 1\\
Vậy\,x = - 1\\
9){x^2} - 9x + 14 = 0\\
\Leftrightarrow {x^2} - 2x - 7x + 14 = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {x - 7} \right) = 0\\
\Leftrightarrow x = 2;x = 7\\
Vậy\,x = 2;x = 7\\
10)\\
6{x^2} - 11x - 35 = 0\\
\Leftrightarrow 6{x^2} - 21x + 10x - 35 = 0\\
\Leftrightarrow 3x\left( {2x - 7} \right) + 5.\left( {2x - 7} \right) = 0\\
\Leftrightarrow \left( {2x - 7} \right)\left( {3x + 5} \right) = 0\\
\Leftrightarrow x = \frac{7}{2};x = \frac{{ - 5}}{3}\\
Vậy\,x = \frac{7}{2};x = \frac{{ - 5}}{3}
\end{array}$
