Đáp án:
e) x=7
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \pm 1\\
A = \dfrac{{2\left( {x + 1} \right) - 3\left( {x - 1} \right) + {x^2} + 7x}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{2x + 2 - 3x + 3 + {x^2} + 7x}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{{x^2} + 6x + 5}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{\left( {x + 5} \right)\left( {x + 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{x + 5}}{{x - 1}}\\
b)A > 0\\
\to \dfrac{{x + 5}}{{x - 1}} > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 5 > 0\\
x - 1 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 5 < 0\\
x - 1 < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
x > 1\\
x < - 5
\end{array} \right.\\
c)A < 1\\
\to \dfrac{{x + 5}}{{x - 1}} < 1\\
\to \dfrac{{x + 5 - x + 1}}{{x - 1}} < 0\\
\to \dfrac{6}{{x - 1}} < 0\\
\to x - 1 < 0\\
\to x < 1\\
d)A = \dfrac{{x + 5}}{{x - 1}} = \dfrac{{x - 1 + 6}}{{x - 1}}\\
= 1 + \dfrac{6}{{x - 1}}\\
A \in Z \to \dfrac{6}{{x - 1}} \in Z\\
\to x - 1 \in U\left( 6 \right)\\
\to \left[ \begin{array}{l}
x - 1 = 6\\
x - 1 = - 6\\
x - 1 = 3\\
x - 1 = - 3\\
x - 1 = 2\\
x - 1 = - 2\\
x - 1 = 1\\
x - 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 7\\
x = - 5\\
x = 4\\
x = - 2\\
x = 3\\
x = - 1(l)\\
x = 2\\
x = 0
\end{array} \right.\\
e)A = 2\\
\to \dfrac{{x + 5}}{{x - 1}} = 2\\
\to x + 5 = 2x - 2\\
\to x = 7
\end{array}\)
