Đáp án:
\(\begin{array}{l}
a)x \ne \left\{ { - 1;0;3} \right\}\\
b)A = \dfrac{{x - 3}}{{x + 1}}\\
c)A = 3\\
d)x = 7
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ { - 1;0;3} \right\}\\
b)A = \dfrac{{2{x^2} + \left( {x - 1} \right)\left( {x + 1} \right) - 3{x^2} + 2x + 1}}{{x\left( {x + 1} \right)}}:\dfrac{{ - x + 5 + x - 3}}{{x - 3}}\\
= \dfrac{{ - {x^2} + {x^2} - 1 + 2x + 1}}{{x\left( {x + 1} \right)}}.\dfrac{{x - 3}}{2}\\
= \dfrac{{2x}}{{x\left( {x + 1} \right)}}.\dfrac{{x - 3}}{2}\\
= \dfrac{{x - 3}}{{x + 1}}\\
c)\left| {x + 2} \right| = 1\\
\to \left[ \begin{array}{l}
x + 2 = 1\\
x + 2 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\left( l \right)\\
x = - 3
\end{array} \right.\\
Thay:x = - 1\\
\to A = \dfrac{{ - 3 - 3}}{{ - 3 + 1}} = \dfrac{6}{2} = 3\\
d)A = \dfrac{1}{2}\\
\to \dfrac{{x - 3}}{{x + 1}} = \dfrac{1}{2}\\
\to 2x - 6 = x + 1\\
\to x = 7
\end{array}\)
