Đáp án:
\(\begin{array}{l}
a) - \dfrac{5}{{21}}\\
b)\dfrac{2}{{\left( {x + 1} \right)\left( {x - 1} \right)}}\\
c)\dfrac{1}{2} > x;x \ne - 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:x = \sqrt {64} = 8\\
\to A = \dfrac{{1 - 2.8}}{{{8^2} - 1}} = - \dfrac{5}{{21}}\\
b)B = \dfrac{{x + 1 + 2\left( {1 - x} \right) - 5 + x}}{{\left( {x + 1} \right)\left( {1 - x} \right)}}\\
= \dfrac{{2x - 4 + 2 - 2x}}{{\left( {x + 1} \right)\left( {1 - x} \right)}}\\
= \dfrac{{ - 2}}{{\left( {x + 1} \right)\left( {1 - x} \right)}} = \dfrac{2}{{\left( {x + 1} \right)\left( {x - 1} \right)}}\\
c)B:A > 0\\
\to \dfrac{2}{{\left( {x + 1} \right)\left( {x - 1} \right)}}:\dfrac{{1 - 2x}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} > 0\\
\to \dfrac{2}{{\left( {x + 1} \right)\left( {x - 1} \right)}}.\dfrac{{\left( {x + 1} \right)\left( {x - 1} \right)}}{{1 - 2x}} > 0\\
\to \dfrac{2}{{1 - 2x}} > 0\\
\to 1 - 2x > 0\\
\to \dfrac{1}{2} > x;x \ne - 1
\end{array}\)
