Đáp án:
d. \( - 1 < x < 0\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ne \pm 1\\
b.M = \left[ {\dfrac{{{{\left( {x + 1} \right)}^2} - {{\left( {1 - x} \right)}^2} + 4{x^2}}}{{\left( {1 - x} \right)\left( {1 + x} \right)}}} \right]:\dfrac{{4\left( {x - 1} \right)\left( {x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}}\\
= \left[ {\dfrac{{{x^2} + 2x + 1 - 1 + 2x - {x^2} + 4{x^2}}}{{ - \left( {x - 1} \right)\left( {x + 1} \right)}}} \right].\dfrac{{x - 1}}{{4\left( {x + 1} \right)}}\\
= \dfrac{{4{x^2} + 4x}}{{ - \left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{x - 1}}{{4\left( {x + 1} \right)}}\\
= \dfrac{{4x\left( {x + 1} \right)}}{{ - \left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{x - 1}}{{4\left( {x + 1} \right)}}\\
= \dfrac{{ - x}}{{x + 1}}\\
c.M = 0\\
\to \dfrac{{ - x}}{{x + 1}} = 0\\
\to x = 0\\
d.M > 0\\
\to \dfrac{{ - x}}{{x + 1}} > 0\\
\to \dfrac{x}{{x + 1}} < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 0\\
x + 1 < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x < 0\\
x + 1 > 0
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 0\\
x < - 1
\end{array} \right.\left( l \right)\\
\left\{ \begin{array}{l}
x < 0\\
x > - 1
\end{array} \right.
\end{array} \right.\\
KL: - 1 < x < 0
\end{array}\)
