Đáp án:
e) \(\left[ \begin{array}{l}
x = - 3\\
x = 0\\
x = - 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\left| {x + 1} \right| = 2\\
\to \left[ \begin{array}{l}
x + 1 = 2\\
x + 1 = - 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\left( l \right)\\
x = - 3
\end{array} \right.\\
Thay:x = - 3\\
\to A = \dfrac{{2.\left( { - 3} \right)}}{{ - 3 + 1}} = 3\\
b)A = - \dfrac{3}{5}\\
\to \dfrac{{2x}}{{x + 1}} = - \dfrac{3}{5}\\
\to 10x = - 3x - 3\\
\to 13x = - 3\\
\to x = - \dfrac{3}{{13}}\\
c)A = \dfrac{{3x}}{{x + 2}}\\
\to \dfrac{{2x}}{{x + 1}} = \dfrac{{3x}}{{x + 2}}\\
\to 2{x^2} + 4x = 3{x^2} + 3x\\
\to {x^2} - x = 0\\
\to x\left( {x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\left( {TM} \right)\\
x = 1\left( l \right)
\end{array} \right.\\
d)P = A - B = \dfrac{{3x}}{{x + 2}} - \left( {\dfrac{{3x}}{{x - 1}} - \dfrac{{{x^2}}}{{{x^2} - 1}}} \right)\\
= \dfrac{{3x}}{{x + 2}} - \dfrac{{3{x^2} + 3x - {x^2}}}{{{x^2} - 1}}\\
= \dfrac{{3x}}{{x + 2}} - \dfrac{{2{x^2} + 3x}}{{{x^2} - 1}}\\
= \dfrac{{3{x^3} - 3x - \left( {x + 2} \right)\left( {2{x^2} + 3x} \right)}}{{\left( {x + 2} \right)\left( {{x^2} - 1} \right)}}\\
= \dfrac{{3{x^3} - 3x - 2{x^3} - 3{x^2} - 4{x^2} - 6x}}{{\left( {x + 2} \right)\left( {{x^2} - 1} \right)}}\\
= \dfrac{{{x^3} - 7{x^2} - 9x}}{{\left( {x + 2} \right)\left( {{x^2} - 1} \right)}}\\
e)A = \dfrac{{2x}}{{x + 1}} = \dfrac{{2\left( {x + 1} \right) - 2}}{{x + 1}} = 2 - \dfrac{2}{{x + 1}}\\
A \in Z\\
\Leftrightarrow \dfrac{2}{{x + 1}} \in Z\\
\Leftrightarrow x + 1 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
x + 1 = 2\\
x + 1 = - 2\\
x + 1 = 1\\
x + 1 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\left( l \right)\\
x = - 3\\
x = 0\\
x = - 2
\end{array} \right.
\end{array}\)
