Đáp án:
B3:
a. \(\left[ \begin{array}{l}
x = 1\\
x = 0\\
x = - 2\\
x = - 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
B3:\\
a.DK:x \ne - 1\\
A = \dfrac{{x + 1 - 2}}{{x + 1}} = 1 - \dfrac{2}{{x + 1}}\\
A \in Z\\
\to \dfrac{2}{{x + 1}} \in Z\\
\to x + 1 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
x + 1 = 2\\
x + 1 = 1\\
x + 1 = - 1\\
x + 1 = - 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = 0\\
x = - 2\\
x = - 3
\end{array} \right.\\
b.DK:x \ne - 1\\
B = \dfrac{{{x^2} + x + x + 1 - 2}}{{x + 1}}\\
= \dfrac{{x\left( {x + 1} \right) + \left( {x + 1} \right) - 2}}{{x + 1}}\\
= \dfrac{{\left( {x + 1} \right)\left( {x + 1} \right) - 2}}{{x + 1}}\\
= x + 1 - \dfrac{2}{{x + 1}}\\
B \in Z\\
\to \dfrac{2}{{x + 1}} \in Z\\
\to \left[ \begin{array}{l}
x = 1\\
x = 0\\
x = - 2\\
x = - 3
\end{array} \right.\\
B4:\\
a.DK:x \ne \left\{ { - 2;1} \right\}\\
A = \dfrac{{3\left( {x - 1} \right) + 2}}{{x - 1}} = 3 + \dfrac{2}{{x - 1}}\\
A \in Z\\
\to \dfrac{2}{{x - 1}} \in Z\\
\to x - 1 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
x - 1 = 2\\
x - 1 = - 2\\
x - 1 = 1\\
x - 1 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = - 1\\
x = 2\\
x = 0
\end{array} \right. \to \left[ \begin{array}{l}
A = 4\\
A = 2\\
A = 5\\
A = 1
\end{array} \right.\\
B = \dfrac{{2{x^2} + x - 1}}{{x + 2}} = \dfrac{{2{x^2} + 4x - 3x - 6 + 5}}{{x + 2}}\\
= \dfrac{{2x\left( {x + 2} \right) - 3\left( {x + 2} \right) + 5}}{{x + 2}}\\
= \dfrac{{\left( {x + 2} \right)\left( {2x - 3} \right) + 5}}{{x + 2}}\\
= 2x - 3 + \dfrac{5}{{x + 2}}\\
B \in Z\\
\to \dfrac{5}{{x + 2}} \in Z\\
\to x + 2 \in U\left( 5 \right)\\
\to \left[ \begin{array}{l}
x + 2 = 5\\
x + 2 = - 5\\
x + 2 = 1\\
x + 2 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = - 7\\
x = - 1\\
x = - 3
\end{array} \right. \to \left[ \begin{array}{l}
B = 4\\
B = - 18\\
B = 0\\
B = - 14
\end{array} \right.
\end{array}\)
b. Để A và B cùng số nguyên
\(\begin{array}{l}
\to A = B = 4\\
\to x = 3
\end{array}\)
