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