Đáp án:
e) \(x < 2;x \ne \left\{ { - 5;1} \right\}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)B = \dfrac{{9 - 3x + {{\left( {x + 5} \right)}^2} - \left( {x + 1} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 5} \right)}}:\dfrac{{7\left( {x - 2} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
= \dfrac{{9 - 3x + {x^2} + 10x + 25 - {x^2} + 1}}{{\left( {x - 1} \right)\left( {x + 5} \right)}}.\dfrac{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}{{7\left( {x - 2} \right)}}\\
= \dfrac{{7x + 35}}{{\left( {x - 1} \right)\left( {x + 5} \right)}}.\dfrac{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}{{7\left( {x - 2} \right)}}\\
= \dfrac{{7\left( {x + 5} \right)}}{{\left( {x - 1} \right)\left( {x + 5} \right)}}.\dfrac{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}{{7\left( {x - 2} \right)}}\\
= \dfrac{{{x^2} + x + 1}}{{x - 2}}\\
b){\left( {x + 5} \right)^2} - 9x - 45 = 0\\
\to {\left( {x + 5} \right)^2} - 9\left( {x + 5} \right) = 0\\
\to \left( {x + 5} \right)\left( {x + 5 - 9} \right) = 0\\
\to \left[ \begin{array}{l}
x = - 5\left( l \right)\\
x = 4
\end{array} \right.\\
Thay:x = 4\\
\to B = \dfrac{{{4^2} + 4 + 1}}{{4 - 2}} = \dfrac{{21}}{2}\\
c)B = \dfrac{{{x^2} + x + 1}}{{x - 2}} = \dfrac{{{x^2} - 4x + 4 + 5x - 3}}{{x - 2}}\\
= \dfrac{{{{\left( {x - 2} \right)}^2} + 5\left( {x - 2} \right) + 7}}{{x - 2}}\\
= \left( {x - 2} \right) + 5 + \dfrac{7}{{x - 2}}\\
B \in Z \Leftrightarrow \dfrac{7}{{x - 2}} \in Z\\
\to x - 2 \in U\left( 7 \right)\\
\to \left[ \begin{array}{l}
x - 2 = 7\\
x - 2 = - 7\\
x - 2 = 1\\
x - 2 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 9\\
x = - 5\\
x = 3\\
x = 1
\end{array} \right.\\
d)B = - \dfrac{3}{4}\\
\to \dfrac{{{x^2} + x + 1}}{{x - 2}} = - \dfrac{3}{4}\\
\to 4{x^2} + 4x + 4 = - 3x + 6\\
\to 4{x^2} + 7x - 2 = 0\\
\to \left( {4x - 1} \right)\left( {x + 2} \right) = 0\\
\to \left[ \begin{array}{l}
x = - 2\\
x = \dfrac{1}{2}
\end{array} \right.\\
e)B < 0\\
\to \dfrac{{{x^2} + x + 1}}{{x - 2}} < 0\\
\to x - 2 < 0\left( {do:{x^2} + x + 1 > 0\forall x} \right)\\
\to x < 2;x \ne \left\{ { - 5;1} \right\}
\end{array}\)
