Đáp án:
B4: c) \(\left[ \begin{array}{l}
x = 5\\
x = 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
B3:\\
a)DK:x \ne \left\{ {0;1;2} \right\}\\
b)P = \dfrac{{x - x + 1}}{{x\left( {x - 1} \right)}}:\dfrac{{\left( {x + 1} \right)\left( {x - 1} \right) - \left( {x + 2} \right)\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {x - 2} \right)}}\\
= \dfrac{1}{{x\left( {x - 1} \right)}}.\dfrac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{{{x^2} - 1 - {x^2} + 4}}\\
= \dfrac{1}{{x\left( {x - 1} \right)}}.\dfrac{{\left( {x - 1} \right)\left( {x - 2} \right)}}{3}\\
= \dfrac{{x - 2}}{{3x}}\\
c)P < 0\\
\to \dfrac{{x - 2}}{{3x}} < 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 2 > 0\\
3x < 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 2 < 0\\
3x > 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > 2\\
x < 0
\end{array} \right.\left( l \right)\\
\left\{ \begin{array}{l}
x < 2\\
x > 0
\end{array} \right.\left( {TM} \right)
\end{array} \right.\\
\to 0 < x < 2;x \ne 1\\
B4:\\
a)DK:x \ne \pm 4\\
B = \dfrac{{x\left( {x - 4} \right) + 4\left( {x + 4} \right)}}{{\left( {x - 4} \right)\left( {x + 4} \right)}}.\dfrac{{x + 4}}{{{x^2} + 16}}\\
= \dfrac{{{x^2} - 4x + 4x + 16}}{{\left( {x - 4} \right)\left( {x + 4} \right)}}.\dfrac{{x + 4}}{{{x^2} + 16}}\\
= \dfrac{{{x^2} + 16}}{{\left( {x - 4} \right)\left( {x + 4} \right)}}.\dfrac{{x + 4}}{{{x^2} + 16}}\\
= \dfrac{1}{{x - 4}}\\
c)B > - 3\\
\to \dfrac{1}{{x - 4}} > - 3\\
\to \dfrac{{1 + 3x - 12}}{{x - 4}} > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
3x - 11 > 0\\
x - 4 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
3x - 11 < 0\\
x - 4 < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
x > 4\\
x < \dfrac{{11}}{3}
\end{array} \right.\\
c)B = \dfrac{1}{{x - 4}}\\
B \in Z \to \dfrac{1}{{x - 4}} \in Z\\
\to x - 4 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
x - 4 = 1\\
x - 4 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 5\\
x = 3
\end{array} \right.
\end{array}\)
