Đáp án:
c) 0<x<1
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:x = - \dfrac{2}{3}\\
\to A = \dfrac{{2 - \dfrac{2}{3}}}{{ - \dfrac{2}{3}}} = - 2\\
b)DK:x \ne \left\{ { - 1;0} \right\}\\
B = \dfrac{{x - 1}}{x} + \dfrac{{2x + 1}}{{{x^2} + x}}\\
= \dfrac{{\left( {x - 1} \right)\left( {x + 1} \right) + 2x + 1}}{{x\left( {x + 1} \right)}}\\
= \dfrac{{{x^2} - 1 + 2x + 1}}{{x\left( {x + 1} \right)}}\\
= \dfrac{{{x^2} + 2x}}{{x\left( {x + 1} \right)}} = \dfrac{{x + 2}}{{x + 1}}\\
c)A:B > \dfrac{3}{2}\\
\to \dfrac{{x + 2}}{x}:\dfrac{{x + 2}}{{x + 1}} > \dfrac{3}{2}\\
\to \dfrac{{x + 2}}{x}.\dfrac{{x + 1}}{{x + 2}} > \dfrac{3}{2}\\
\to \dfrac{{x + 1}}{x} > \dfrac{3}{2}\\
\to \dfrac{{2x + 1 - 3x}}{{2x}} > 0\\
\to \dfrac{{1 - x}}{{2x}} > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
1 - x > 0\\
x > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
1 - x < 0\\
x < 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
1 > x\\
x > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
1 < x\\
x < 0
\end{array} \right.\left( l \right)
\end{array} \right.\\
\to 0 < x < 1
\end{array}\)
