Đáp án:
\(\begin{array}{l}
B23:\\
a) - \dfrac{1}{{x - 2}}\\
b)\left[ \begin{array}{l}
C = \dfrac{2}{3}\\
C = \dfrac{2}{5}
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B22:\\
a)DK:x \ne \left\{ { - 2;0;2} \right\}\\
B = \dfrac{{4x\left( {x - 2} \right) - 8{x^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}:\dfrac{{x - 1 - 2\left( {x - 2} \right)}}{{x\left( {x - 2} \right)}}\\
= \dfrac{{ - 4{x^2} - 8x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{x\left( {x - 2} \right)}}{{3 - x}}\\
= \dfrac{{ - 4x\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{x\left( {x - 2} \right)}}{{3 - x}}\\
= \dfrac{{4{x^2}}}{{x - 3}}\\
b)B = 1\\
\to \dfrac{{4{x^2}}}{{x - 3}} = 1\\
\to 4{x^2} = x - 3\\
\to 4{x^2} - x + 3 = 0\left( {vô nghiệm} \right)\\
\to x \in \emptyset \\
B23:\\
a)DK:x \ne \left\{ { - 2;0;2} \right\}\\
C = \left[ {\dfrac{{{x^2}}}{{x\left( {x - 2} \right)\left( {x + 2} \right)}} - \dfrac{{3.2}}{{3\left( {x - 2} \right)}} + \dfrac{1}{{x + 2}}} \right]:\dfrac{{\left( {x - 2} \right)\left( {x + 2} \right) + 10 - {x^2}}}{{x + 2}}\\
= \dfrac{{x - 2\left( {x + 2} \right) + x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{x + 2}}{{{x^2} - 4 + 10 - {x^2}}}\\
= \dfrac{{ - 6}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{x + 2}}{6}\\
= - \dfrac{1}{{x - 2}}\\
b)\left| x \right| = \dfrac{1}{2}\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = - \dfrac{1}{2}
\end{array} \right.\\
Thay:\left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = - \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
C = - \dfrac{1}{{\dfrac{1}{2} - 2}}\\
C = - \dfrac{1}{{ - \dfrac{1}{2} - 2}}
\end{array} \right. \to \left[ \begin{array}{l}
C = \dfrac{2}{3}\\
C = \dfrac{2}{5}
\end{array} \right.
\end{array}\)
\(\begin{array}{l}
B24:\\
DK:x \ne - 1\\
a)D = 1 + \left( {\dfrac{{x + 1 + x + 1 - 2\left( {{x^2} - x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}} \right):\dfrac{{{x^2}\left( {x - 2} \right)}}{{x\left( {{x^2} - x + 1} \right)}}\\
= 1 + \dfrac{{ - 2{x^2} + 4x}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}.\dfrac{{x\left( {{x^2} - x + 1} \right)}}{{{x^2}\left( {x - 2} \right)}}\\
= 1 + \dfrac{{ - 2x\left( {x - 2} \right)}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}.\dfrac{{x\left( {{x^2} - x + 1} \right)}}{{{x^2}\left( {x - 2} \right)}}\\
= 1 - \dfrac{2}{{x + 1}} = \dfrac{{x + 1 - 2}}{{x + 1}}\\
= \dfrac{{x - 1}}{{x + 1}}\\
b)\left| {x - \dfrac{3}{4}} \right| = \dfrac{5}{4}\\
\to \left[ \begin{array}{l}
x - \dfrac{3}{4} = \dfrac{5}{4}\\
x - \dfrac{3}{4} = - \dfrac{5}{4}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = - \dfrac{1}{2}
\end{array} \right.
\end{array}\)
\(\begin{array}{l}
B24:\\
DK:x \ne - 1\\
a)D = 1 + \left( {\dfrac{{x + 1 + x + 1 - 2\left( {{x^2} - x + 1} \right)}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}} \right):\dfrac{{{x^2}\left( {x - 2} \right)}}{{x\left( {{x^2} - x + 1} \right)}}\\
= 1 + \dfrac{{ - 2{x^2} + 4x}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}.\dfrac{{x\left( {{x^2} - x + 1} \right)}}{{{x^2}\left( {x - 2} \right)}}\\
= 1 + \dfrac{{ - 2x\left( {x - 2} \right)}}{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)}}.\dfrac{{x\left( {{x^2} - x + 1} \right)}}{{{x^2}\left( {x - 2} \right)}}\\
= 1 - \dfrac{2}{{x + 1}} = \dfrac{{x + 1 - 2}}{{x + 1}}\\
= \dfrac{{x - 1}}{{x + 1}}\\
b)\left| {x - \dfrac{3}{4}} \right| = \dfrac{5}{4}\\
\to \left[ \begin{array}{l}
x - \dfrac{3}{4} = \dfrac{5}{4}\\
x - \dfrac{3}{4} = - \dfrac{5}{4}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = - \dfrac{1}{2}
\end{array} \right.
\end{array}\)
( bạn xem lại đề câu 21 nha, có thiếu dấu hoặc số không nhé )
