Đáp án:
$\begin{array}{l}
5a)3{\left( {x - 1} \right)^2} - 3x\left( {x - 5} \right) = 1\\
\Leftrightarrow 3\left( {{x^2} - 2x + 1} \right) - 3{x^2} + 15x = 1\\
\Leftrightarrow 3{x^2} - 6x + 3 - 3{x^2} + 15x = 1\\
\Leftrightarrow 9x = - 2\\
\Leftrightarrow x = - \dfrac{2}{9}\\
Vậy\,x = - \dfrac{2}{9}\\
b){\left( {6x - 2} \right)^2} + {\left( {5x - 2} \right)^2} - 4\left( {3x - 1} \right)\left( {5x - 2} \right) = 0\\
\Leftrightarrow {\left( {6x - 2} \right)^2} - 2.\left( {6x - 2} \right)\left( {5x - 2} \right) + {\left( {5x - 2} \right)^2} = 0\\
\Leftrightarrow {\left( {6x - 2 - 5x + 2} \right)^2} = 0\\
\Leftrightarrow {x^2} = 0\\
\Leftrightarrow x = 0\\
Vậy\,x = 0\\
B6)\\
a)4{x^2} - 6x + \dfrac{9}{4}\\
= {\left( {2x} \right)^2} - 2.2x.\dfrac{3}{2} + {\left( {\dfrac{3}{2}} \right)^2}\\
= {\left( {2x - \dfrac{3}{2}} \right)^2}\\
b)4\left( {{x^2} + 2x + 1} \right) - 12x - 3\\
= 4{x^2} + 8x + 4 - 12x - 3\\
= 4{x^2} - 4x + 1\\
= {\left( {2x - 1} \right)^2}\\
c)25{x^2} - 20xy + 4{y^2}\\
= {\left( {5x - 2y} \right)^2}
\end{array}$
