Đáp án:
\(\left[ \begin{array}{l}
x > 1\\
x < 0;x \ne - 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
c)B = \dfrac{{{x^2} + 2x + 1 + {x^2} - 3x + 2 - 2{x^2} - x - 5}}{{\left( {x + 1} \right)\left( {x - 1} \right)}}\\
= \dfrac{{ - 2x - 2}}{{\left( {x + 1} \right)\left( {x - 1} \right)}} = - \dfrac{2}{{x - 1}}\\
B < 2\\
\to - \dfrac{2}{{x - 1}} < 2\\
\to \dfrac{{ - 2 - x + 2}}{{x - 1}} < 0\\
\to \dfrac{{ - x}}{{x - 1}} < 0\\
\to \dfrac{x}{{x - 1}} > 0\\
\to \left[ \begin{array}{l}
x > 1\\
x < 0;x \ne - 1
\end{array} \right.
\end{array}\)
