Đáp án:
\(\begin{array}{l}
1)a)C = 2024\\
b)\left[ \begin{array}{l}
a = 5\\
a = - 3
\end{array} \right.\\
2)Q = \dfrac{{3x - 5}}{{2x + 1}}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)a)C = \left[ {\dfrac{{\left( {a + 1} \right)\left( {{a^2} - a + 1} \right)}}{{\left( {a - 2} \right)\left( {a + 1} \right)}} + \dfrac{{{a^2}}}{{a\left( {2 - a} \right)}}} \right].\dfrac{{2 - a}}{{1 - a}}\\
= \left[ {\dfrac{{{a^2} - a + 1}}{{a - 2}} - \dfrac{a}{{a - 2}}} \right].\dfrac{{a - 2}}{{a - 1}}\\
= \dfrac{{{a^2} - 2a + 1}}{{a - 2}}.\dfrac{{a - 2}}{{a - 1}}\\
= \dfrac{{{{\left( {a - 1} \right)}^2}}}{{a - 2}}.\dfrac{{a - 2}}{{a - 1}}\\
= a - 1\\
Thay:a = 2025\\
\to C = 2025 - 1 = 2024\\
b)\left| C \right| = 4\\
\to \left[ \begin{array}{l}
C = 4\\
C = - 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
a - 1 = 4\\
a - 1 = - 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
a = 5\\
a = - 3
\end{array} \right.\\
2)Q = \dfrac{{4x - 6 - x + 1}}{{2x - 3}}:\dfrac{{6x + 1 + x\left( {2x - 3} \right)}}{{\left( {x + 1} \right)\left( {2x - 3} \right)}}\\
= \dfrac{{3x - 5}}{{2x - 3}}.\dfrac{{\left( {x + 1} \right)\left( {2x - 3} \right)}}{{2{x^2} + 3x + 1}}\\
= \dfrac{{3x - 5}}{{2x - 3}}.\dfrac{{\left( {x + 1} \right)\left( {2x - 3} \right)}}{{\left( {2x + 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{3x - 5}}{{2x + 1}}
\end{array}\)
