Đáp án:
a) \( - \dfrac{{3x - 4}}{4}\)
b) \(\left[ \begin{array}{l}
x = - \dfrac{{16}}{{11}}\\
x = - \dfrac{{24}}{5}
\end{array} \right.\)
c) \(\left[ \begin{array}{l}
x = - \dfrac{{16}}{{11}}\\
x = - \dfrac{{24}}{5}
\end{array} \right.\)
d) \(Min = - \dfrac{1}{{12}}\)
g) \(\dfrac{{12}}{5} > x \ge \dfrac{4}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ {0;2} \right\}\\
B = \dfrac{{x - 4 + 2x}}{{x\left( {x - 2} \right)}}:\dfrac{{{x^2} - 4 - {x^2}}}{{x\left( {x - 2} \right)}}\\
= \dfrac{{3x - 4}}{{x\left( {x - 2} \right)}}.\dfrac{{x\left( {x - 2} \right)}}{{ - 4}}\\
= - \dfrac{{3x - 4}}{4}\\
b)Thay:x = - 2\\
\to B = - \dfrac{{3.\left( { - 2} \right) - 4}}{4} = \dfrac{5}{2}\\
c)\left| B \right| = 2x + 5\\
\to \left[ \begin{array}{l}
B = 2x + 5\\
B = - 2x - 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
- \dfrac{{3x - 4}}{4} = 2x + 5\\
- \dfrac{{3x - 4}}{4} = - 2x - 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
- 3x + 4 = 8x + 20\\
3x - 4 = 8x + 20
\end{array} \right.\\
\to \left[ \begin{array}{l}
11x = - 16\\
5x = - 24
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{{16}}{{11}}\\
x = - \dfrac{{24}}{5}
\end{array} \right.\\
d)\left( {2 - x} \right).B = \left( {x - 2} \right).\dfrac{{3x - 4}}{4}\\
= \dfrac{{3{x^2} - 10x + 8}}{4}\\
= \dfrac{{3{x^2} - 2.x\sqrt 3 .\dfrac{5}{{\sqrt 3 }} + \dfrac{{25}}{3} - \dfrac{1}{3}}}{4}\\
= \dfrac{{{{\left( {x\sqrt 3 - \dfrac{5}{{\sqrt 3 }}} \right)}^2} - \dfrac{1}{3}}}{4}\\
Do:{\left( {x\sqrt 3 - \dfrac{5}{{\sqrt 3 }}} \right)^2} \ge 0\forall x\\
\to {\left( {x\sqrt 3 - \dfrac{5}{{\sqrt 3 }}} \right)^2} - \dfrac{1}{3} \ge - \dfrac{1}{3}\\
\to \dfrac{{{{\left( {x\sqrt 3 - \dfrac{5}{{\sqrt 3 }}} \right)}^2} - \dfrac{1}{3}}}{4} \ge - \dfrac{1}{{12}}\\
\to Min = - \dfrac{1}{{12}}\\
\Leftrightarrow x\sqrt 3 - \dfrac{5}{{\sqrt 3 }} = 0\\
\to x = \dfrac{5}{3}\\
g)\left| B \right| + 3 < 2x - 1\\
\to \left[ \begin{array}{l}
- \dfrac{{3x - 4}}{4} + 3 < 2x - 1\left( {DK:x < \dfrac{4}{3}} \right)\\
\dfrac{{3x - 4}}{4} + 3 > 2x - 1\left( {DK:x \ge \dfrac{4}{3}} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
- 3x + 4 + 12 < 8x - 4\\
3x - 4 + 12 > 8x - 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
11x > 20\\
5x < 12
\end{array} \right.\\
\to \left[ \begin{array}{l}
x > \dfrac{{20}}{{11}}\left( l \right)\\
x < \dfrac{{12}}{5}
\end{array} \right.\\
\to \dfrac{{12}}{5} > x \ge \dfrac{4}{3}
\end{array}\)
