Đáp án:
B4:
c) \(\left[ \begin{array}{l}
x = 4\\
x = 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
B3:\\
a)Thay:x = - 2\\
\to B = \dfrac{{2.\left( { - 2} \right) + 1}}{{{{\left( { - 2} \right)}^2} - 1}} = \dfrac{{ - 3}}{3} = - 1\\
b)A = \dfrac{{3x + 1 - x\left( {x + 1} \right) + {{\left( {x - 1} \right)}^2}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{3x + 1 - {x^2} - x + {x^2} - 2x + 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{2}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
c)P = A:B = \dfrac{2}{{\left( {x - 1} \right)\left( {x + 1} \right)}}:\dfrac{{2x + 1}}{{{x^2} - 1}}\\
= \dfrac{2}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{\left( {x - 1} \right)\left( {x + 1} \right)}}{{2x + 1}}\\
= \dfrac{2}{{2x + 1}}\\
P = 3 \to \dfrac{2}{{2x + 1}} = 3\\
\to 6x + 3 = 2\\
\to 6x = - 1\\
\to x = - \dfrac{1}{6}\\
B4:\\
a)A = \left[ {\dfrac{{2x - 3 - 2\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}} \right].\dfrac{{x + 3}}{3}\\
= \dfrac{{2x - 3 - 2x + 6}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\dfrac{{x + 3}}{3}\\
= \dfrac{3}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\dfrac{{x + 3}}{3}\\
= \dfrac{1}{{x - 3}}\\
b)Thay:x = - \dfrac{1}{2}\\
\to A = \dfrac{1}{{ - \dfrac{1}{2} - 3}} = - \dfrac{2}{7}\\
c)A \in Z \to \dfrac{1}{{x - 3}} \in Z\\
\to x - 3 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
x - 3 = 1\\
x - 3 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 4\\
x = 2
\end{array} \right.
\end{array}\)
