Đáp án:
$\begin{array}{l}
1)\frac{{{x^2} + x - 2}}{{x + 2}} = 10\left( {dkxd:x \ne - 2} \right)\\
\Rightarrow {x^2} + x - 2 = 10x + 20\\
\Rightarrow {x^2} - 9x - 22 = 0\\
\Rightarrow {x^2} - 11x + 2x - 22 = 0\\
\Rightarrow \left( {x - 11} \right)\left( {x + 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 11\left( {tm} \right)\\
x = - 2\left( {ktm} \right)
\end{array} \right.\\
Vay\,x = 11\\
2)x - 1 + \frac{2}{{x - 2}} = \frac{{2x - 2}}{{x - 2}}\left( {dkxd:x \ne 2} \right)\\
\Leftrightarrow \frac{{\left( {x - 1} \right)\left( {x - 2} \right) + 2}}{{x - 2}} = \frac{{2x - 2}}{{x - 2}}\\
\Rightarrow {x^2} - 3x + 2 + 2 = 2x - 2\\
\Rightarrow {x^2} - 5x + 6 = 0\\
\Rightarrow \left( {x - 3} \right)\left( {x - 2} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 3\left( {tm} \right)\\
x = 2\left( {ktm} \right)
\end{array} \right.\\
Vay\,x = 3
\end{array}$
Em chia nhỏ câu hỏi ra nhé
