Đáp án:
$\begin{array}{l}
a){\left( {x - 2} \right)^2} - \left( {x - 3} \right)\left( {x + 3} \right) = 0\\
\Leftrightarrow {x^2} - 4x + 4 - \left( {{x^2} - 9} \right) = 0\\
\Leftrightarrow {x^2} - 4x + 4 - {x^2} + 9 = 0\\
\Leftrightarrow 4x = 13\\
\Leftrightarrow x = \dfrac{{13}}{4}\\
Vậy\,x = \dfrac{{13}}{4}\\
b){\left( {3x - 1} \right)^2} - 16 = 0\\
\Leftrightarrow \left( {3x - 1 - 4} \right)\left( {3x - 1 + 4} \right) = 0\\
\Leftrightarrow \left( {3x - 5} \right)\left( {3x + 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
3x = 5\\
3x = - 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{5}{3}\\
x = - 1
\end{array} \right.\\
Vậy\,x = \dfrac{5}{3};x = - 1\\
c)5{x^3} - 35x = 0\\
\Leftrightarrow 5x\left( {{x^2} - 7} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} = 7
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - \sqrt 7 \\
x = \sqrt 7
\end{array} \right.\\
Vậy\,x = 0;x = \sqrt 7 ;x = - \sqrt 7 \\
d){\left( {x - 3} \right)^2} = 4{x^2} - 20x + 25\\
\Leftrightarrow {\left( {x - 3} \right)^2} = {\left( {2x - 5} \right)^2}\\
\Leftrightarrow {\left( {x - 3} \right)^2} - {\left( {2x - 5} \right)^2} = 0\\
\Leftrightarrow \left( {x - 3 - 2x + 5} \right)\left( {x - 3 + 2x - 5} \right) = 0\\
\Leftrightarrow \left( {2 - x} \right)\left( {3x - 8} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = \dfrac{8}{3}
\end{array} \right.\\
Vậy\,x = 2;x = \dfrac{8}{3}\\
e)x + {x^2} - {x^3} - {x^4} = 0\\
\Leftrightarrow x\left( {1 + x} \right) - {x^3}\left( {1 + x} \right) = 0\\
\Leftrightarrow \left( {1 + x} \right).x.\left( {1 - {x^2}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
1 + x = 0\\
x = 0\\
1 - {x^2} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - 1\\
x = 0\\
x = 1
\end{array} \right.\\
Vậy\,x = 0;x = 1;x = - 1\\
f){\left( {7x - 5} \right)^2} - 2\left( {7x - 5} \right)\left( {x + 3} \right) + {\left( {x + 3} \right)^2} = 0\\
\Leftrightarrow {\left( {7x - 5 - x - 3} \right)^2} = 0\\
\Leftrightarrow {\left( {6x - 8} \right)^2} = 0\\
\Leftrightarrow 6x - 8 = 0\\
\Leftrightarrow x = \dfrac{4}{3}\\
Vậy\,x = \dfrac{4}{3}
\end{array}$
