Đáp án:
$\begin{array}{l}
a)\left\{ \begin{array}{l}
3y + 2x = 21\\
10x + 3y = 45
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
10x + 15y = 105\\
10x + 3y = 45
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
10x = 105 - 15y\\
10x + 3y = 45
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
10x = 105 - 15y\\
105 - 15y + 3y = 45
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
2x + 3y = 21\\
12y = 105 - 45 = 60
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
y = 5\\
2x = 21 - 3y
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
y = 5\\
x = 3
\end{array} \right.\\
Vậy\,x = 3;y = 5\\
b)\left\{ \begin{array}{l}
\left( {x + 1} \right)\left( {x - 1} \right) = \left( {x - 2} \right)\left( {y + 1} \right) - 1\\
2\left( {x - 2} \right).y - x = 2x - 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x^2} - 1 = \left( {x - 2} \right).y + x - 2 - 1\\
2.\left( {x - 2} \right).y = 3x - 3
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x^2} = \left( {x - 2} \right).y + x - 2\\
\left( {x - 2} \right).y = \dfrac{{3x - 3}}{2}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x^2} = \dfrac{{3x - 3}}{2} + x - 2\\
\left( {x - 2} \right).y = \dfrac{{3x - 3}}{2}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
2{x^2} - 5x + 7 = 0\left( {vn} \right)\\
y = \dfrac{{3x - 3}}{{2\left( {x - 2} \right)}}
\end{array} \right.
\end{array}$
Vậy hệ pt vô nghiệm
