Đáp án:
\(\begin{array}{l}
a)B = - \dfrac{4}{{15}}\\
b)x > 0;x \ne 2
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Thay:x = 8\\
\to B = \dfrac{{ - 16}}{{{8^2} - 4}} = - \dfrac{{16}}{{60}} = - \dfrac{4}{{15}}\\
b)A = \dfrac{{{{\left( {x + 2} \right)}^2} - {{\left( {x - 2} \right)}^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{{x^2} + 4x + 4 - {x^2} + 4x - 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{8x}}{{{x^2} - 4}}\\
A:B = \dfrac{{8x}}{{{x^2} - 4}}:\dfrac{{ - 16}}{{{x^2} - 4}}\\
= - \dfrac{{8x}}{{16}} = - \dfrac{x}{2}\\
A:B < 0\\
\to - \dfrac{x}{2} < 0\\
\to \dfrac{x}{2} > 0\\
\to x > 0;x \ne 2
\end{array}\)
