Đáp án:
$\begin{array}{l}
4)a)Do:\left( {x - y} \right) - \left( {y - x} \right)\\
= x - y - y + x\\
= 2x - 2y \ne 0\\
\Rightarrow x - y \ne y - x\\
b){\left( {x + 1} \right)^2} = {x^2} + 2x + 1\\
\Rightarrow {\left( {x + 1} \right)^2} \ne {x^2} + 1\\
c){\left( {x - y} \right)^3} = {x^3} - 3{x^2}y + 3x{y^2} - {y^3}\\
{\left( {y - x} \right)^3} = {y^3} - 3{y^2}x + 3y.{x^2} - {x^3}\\
\Rightarrow {\left( {x - y} \right)^3} \ne {\left( {y - x} \right)^3}\\
9)\\
a)3x = \left( {a + b} \right).x + 2a - b\\
\Rightarrow \left\{ \begin{array}{l}
a + b = 3\\
2a - b = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a + b = 3\\
b = 2a
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a + 2a = 3\\
b = 2a
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a = 1\\
b = 2
\end{array} \right.\\
b)\left( {x + a} \right).\left( {b.x - 1} \right) = {x^2} - 7a + 6\\
\Rightarrow \left( {x + a} \right).\left( {b.x - 1} \right) = \left( {x - 1} \right)\left( {x - 6} \right)\\
\Rightarrow \left\{ \begin{array}{l}
x + a = x - 6\\
b.x - 1 = x - 1
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
a = - 6\\
b = 1
\end{array} \right.\\
10)\dfrac{{3y\left( {x + 1} \right) - 6x - 6}}{{3y - 6}}\\
= \dfrac{{3y\left( {x + 1} \right) - 6\left( {x + 1} \right)}}{{3y - 6}}\\
= \dfrac{{\left( {x + 1} \right)\left( {3y - 6} \right)}}{{3y - 6}}\\
= x + 1\\
\dfrac{{2\left( {y + 3} \right) + 2xy + 6x}}{{2y + 6}}\\
= \dfrac{{2y + 6 + x\left( {2y + 6} \right)}}{{2y + 6}}\\
= \dfrac{{\left( {2y + 6} \right)\left( {x + 1} \right)}}{{2y + 6}}\\
= x + 1\\
Vậy\,\dfrac{{3y\left( {x + 1} \right) - 6x - 6}}{{3y - 6}} = \dfrac{{2\left( {y + 3} \right) + 2xy + 6x}}{{2y + 6}}
\end{array}$
