Đáp án:
B8:
a) \( - \dfrac{1}{{{x^2} + x + 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
B7:\\
a)DK:x \ne 2\\
N = \dfrac{{3\left( {{x^2} - 4} \right)}}{{{{\left( {x - 2} \right)}^2}}} - \dfrac{{2\left( {2x - 1} \right)}}{{2\left( {x - 2} \right)}} + \dfrac{{x - 5}}{{x - 2}}\\
= \dfrac{{3\left( {x + 2} \right)}}{{x - 2}} - \dfrac{{2x - 1}}{{x - 2}} + \dfrac{{x - 5}}{{x - 2}}\\
= \dfrac{{3x + 6 - 2x + 1 + x - 5}}{{x - 2}}\\
= \dfrac{{2x + 2}}{{x - 2}}\\
b)\left| {x + 1} \right| = 3\\
\to \left[ \begin{array}{l}
x + 1 = 3\\
x + 1 = - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\left( l \right)\\
x = - 4
\end{array} \right.\\
Thay:x = - 4\\
\to N = \dfrac{{2\left( { - 4} \right) + 2}}{{ - 4 - 2}} = 1\\
c)N = \dfrac{{2x + 2}}{{x - 2}} = \dfrac{{2\left( {x - 2} \right) + 6}}{{x - 2}} = 2 + \dfrac{6}{{x - 2}}\\
N \in Z \Leftrightarrow \dfrac{6}{{x - 2}} \in Z\\
\Leftrightarrow x - 2 \in U\left( 6 \right)\\
\to \left[ \begin{array}{l}
x - 2 = 6\\
x - 2 = - 6\\
x - 2 = 3\\
x - 2 = - 3\\
x - 2 = 2\\
x - 2 = - 2\\
x - 2 = 1\\
x - 2 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 8\\
x = - 4\\
x = 5\\
x = - 1\\
x = 4\\
x = 0\\
x = 3\\
x = 1
\end{array} \right.\\
B8:\\
a)DK:x \ne 1\\
P = \dfrac{{3 + \left( {x + 1} \right)\left( {x - 1} \right) - {x^2} - x - 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \dfrac{{3 + {x^2} - 1 - {x^2} - x - 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \dfrac{{ - x + 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = - \dfrac{1}{{{x^2} + x + 1}}\\
b)\left| {3x - 1} \right| = 2\\
\to \left[ \begin{array}{l}
3x - 1 = 2\\
3x - 1 = - 2
\end{array} \right. \to \left[ \begin{array}{l}
x = 1\left( l \right)\\
x = - \dfrac{1}{3}
\end{array} \right.\\
Thay:x = - \dfrac{1}{3}\\
\to P = - \dfrac{1}{{{{\left( { - \dfrac{1}{3}} \right)}^2} - \dfrac{1}{3} + 1}} = - \dfrac{9}{7}\\
c)Do:{x^2} + x + 1 = {x^2} + 2x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4}\\
= {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} > 0\forall x \ne 1\\
\to - \dfrac{1}{{{x^2} + x + 1}} < 0\forall x \ne 1
\end{array}\)
