Đáp án:
a) \(A = - 2\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:x = - \dfrac{2}{3}\\
\to A = \dfrac{{2 - \dfrac{2}{3}}}{{ - \dfrac{2}{3}}} = - 2\\
b)DK:x \ne \left\{ { - 1;0} \right\}\\
B = \dfrac{{x - 1}}{{2x}} + \dfrac{1}{{{x^2} + x}}\\
= \dfrac{{\left( {x - 1} \right)\left( {x + 1} \right) + 2}}{{2x\left( {x + 1} \right)}}\\
= \dfrac{{{x^2} - 1 + 2}}{{2x\left( {x + 1} \right)}}\\
= \dfrac{{{x^2} + 1}}{{2x\left( {x + 1} \right)}}\\
c)A:B > \dfrac{3}{2}\\
\to \dfrac{{x + 2}}{x}:\dfrac{{{x^2} + 1}}{{2x\left( {x + 1} \right)}} > \dfrac{3}{2}\\
\to \dfrac{{x + 2}}{x}.\dfrac{{2x\left( {x + 1} \right)}}{{{x^2} + 1}} > \dfrac{3}{2}\\
\to \dfrac{{2\left( {x + 1} \right)\left( {x + 2} \right)}}{{{x^2} + 1}} > \dfrac{3}{2}\\
\to \dfrac{{2\left( {2{x^2} + 6x + 4} \right) - 3\left( {{x^2} + 1} \right)}}{{2\left( {{x^2} + 1} \right)}} > 0\\
\to 4{x^2} + 12x + 8 - 3{x^2} - 3 > 0\left( {do:{x^2} + 1 > 0\forall x} \right)\\
\to {x^2} + 12x + 5 > 0\\
\to {x^2} + 12x + 36 - 31 > 0\\
\to {\left( {x + 6} \right)^2} > 31\\
\to \left[ \begin{array}{l}
x + 6 > \sqrt {31} \\
x + 6 < - \sqrt {31}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x > - 6 + \sqrt {31} \\
x < - 6 - \sqrt {31}
\end{array} \right.
\end{array}\)
