Đáp án:
a) \(x \ne \left\{ {0;\dfrac{1}{2}} \right\}\)
Giải thích các bước giải:
\(\begin{array}{l}
B14:\\
a)DK:2{x^2} - x \ne 0 \to x\left( {2x - 1} \right) \ne 0\\
\to x \ne \left\{ {0;\dfrac{1}{2}} \right\}\\
b)A = \dfrac{{x\left( {4{x^2} - 1} \right)}}{{x\left( {2x - 1} \right)}} = \dfrac{{x\left( {2x - 1} \right)\left( {2x + 1} \right)}}{{x\left( {2x - 1} \right)}}\\
= 2x + 1\\
c)Thay:x = - \dfrac{1}{2}\\
\to A = 2.\left( { - \dfrac{1}{2}} \right) + 1 = 0\\
d)A = 0\\
\to 2x + 1 = 0\\
\to x = - \dfrac{1}{2}
\end{array}\)
