Đáp án:
a) \(\left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne - 3\\
\dfrac{{{x^2} - x}}{{x + 3}} = 0\\
\to {x^2} - x = 0\\
\to x\left( {x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x - 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\\
b)DK:x \ne 1\\
\dfrac{{{x^2} - 1}}{{x - 1}} = 0\\
\to \dfrac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{x - 1}} = 0\\
\to x + 1 = 0\\
\to x = - 1\left( {TM} \right)\\
c)DK:x \ne \left\{ {2;3} \right\}\\
\dfrac{{{x^2} - 2x + 2}}{{\left( {x - 2} \right)\left( {x - 3} \right)}} = 0\\
\to {x^2} - 2x + 2 = 0\left( {vô nghiệm} \right)\\
Do:{x^2} - 2x + 1 + 2 = {\left( {x - 1} \right)^2} + 2 > 0\forall x \ne \left\{ {2;3} \right\}\\
\to x \in \emptyset
\end{array}\)
