Đáp án:
d) x=-2
Giải thích các bước giải:
\(\begin{array}{l}
B16:\\
a)DK:x \ne \left\{ { - 3;2;3} \right\}\\
F = \left[ {\dfrac{{x\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} - 1} \right]:\left[ {\dfrac{{9 - {x^2} + \left( {x - 3} \right)\left( {x + 3} \right) - {{\left( {x - 2} \right)}^2}}}{{\left( {x - 2} \right)\left( {x + 3} \right)}}} \right]\\
= \dfrac{{x - x - 3}}{{x + 3}}.\dfrac{{\left( {x - 2} \right)\left( {x + 3} \right)}}{{9 - {x^2} + {x^2} - 9 - {{\left( {x - 2} \right)}^2}}}\\
= \dfrac{{ - 3}}{{x + 3}}.\dfrac{{\left( {x - 2} \right)\left( {x + 3} \right)}}{{ - {{\left( {x - 2} \right)}^2}}}\\
= \dfrac{3}{{x + 3}}.\dfrac{{\left( {x - 2} \right)\left( {x + 3} \right)}}{{{{\left( {x - 2} \right)}^2}}}\\
= \dfrac{3}{{x - 2}}\\
b)Thay:x = \dfrac{2}{5}\\
\to F = \dfrac{3}{{\dfrac{2}{5} - 2}} = - \dfrac{{15}}{8}\\
c)F \in Z\\
\Leftrightarrow \dfrac{3}{{x - 2}} \in Z\\
\Leftrightarrow x - 2 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
x - 2 = 3\\
x - 2 = - 3\\
x - 2 = 1\\
x - 2 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 5\\
x = - 1\\
x = 3\left( l \right)\\
x = 1
\end{array} \right.\\
d)F = - \dfrac{3}{4}\\
\to \dfrac{3}{{x - 2}} = - \dfrac{3}{4}\\
\to - 3x + 6 = 12\\
\to - 3x = 6\\
\to x = - 2
\end{array}\)
