Giải thích các bước giải:
B1:
$\begin{array}{l}
a)6{x^3}{y^2} - \dfrac{2}{3}{x^2}{y^3} = {x^2}y\left( {6xy - \dfrac{2}{3}{y^2}} \right)\\
\Rightarrow \left( {6{x^3}{y^2} - \dfrac{2}{3}{x^2}{y^3}} \right):2{x^2}y = {x^2}y\left( {6xy - \dfrac{2}{3}{y^2}} \right):2{x^2}y = 3xy - \dfrac{1}{3}{y^2}
\end{array}$
Vậy $\left( {6{x^3}{y^2} - \dfrac{2}{3}{x^2}{y^3}} \right):2{x^2}y = 3xy - \dfrac{1}{3}{y^2}$
$\begin{array}{l}
b){(x - 2)^2} - (x - 3)(x - 1)\\
= {x^2} - 4x + 4 - {x^2} + 4x - 3\\
= 1
\end{array}$
Vậy ${(x - 2)^2} - (x - 3)(x - 1) = 1$
B2:
$\begin{array}{l}
a)x(2x + 3) - 2{x^2} = 6\\
\Leftrightarrow 2{x^2} + 3x - 2{x^2} = 6\\
\Leftrightarrow 3x = 6\\
\Leftrightarrow x = 2
\end{array}$
Vậy $x = 2$
$\begin{array}{l}
b){(x + 1)^2} - 4{x^2} = 0\\
\Leftrightarrow \left( {x + 1 - 2x} \right)\left( {x + 1 + 2x} \right) = 0\\
\Leftrightarrow \left( {1 - x} \right)\left( {3x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
1 - x = 0\\
3x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \dfrac{{ - 1}}{3}
\end{array} \right.
\end{array}$
Vậy $x \in \left\{ {\dfrac{{ - 1}}{3};1} \right\}$
