Đáp án:
B1:
d. \(\left[ \begin{array}{l}
x = 4\\
x = \dfrac{1}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a.x = 3 + 5\\
\to x = 8\\
c.DK:x \ne \left\{ {0;1} \right\}\\
\dfrac{{x - 4 - x + 1 + 3x}}{{x\left( {x - 1} \right)}} = 0\\
\to 3x - 3 = 0\\
\to x = 1\left( l \right)\\
\to x \in \emptyset \\
b.3x - 9 \ge x - 17\\
\to 2x \ge - 8\\
\to x \ge - 4\\
d.3x - 5 = \left| {x + 3} \right|\\
\to \left[ \begin{array}{l}
x + 3 = 3x - 5\\
x + 3 = - 3x + 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 8\\
4x = 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 4\\
x = \dfrac{1}{2}
\end{array} \right.
\end{array}\)
