Đáp án:
a) \(x = - \dfrac{{16}}{{11}}\)
b) \(Min = - \dfrac{1}{{12}}\)
c) Không tồn tại x TMĐK
d) \(\dfrac{{12}}{5} > x > \dfrac{4}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ {0;2} \right\}\\
a)B = \dfrac{{x - 4 + 2x}}{{x\left( {x - 2} \right)}}:\dfrac{{{x^2} - 4 - {x^2}}}{{x\left( {x - 2} \right)}}\\
= \dfrac{{3x - 4}}{{x\left( {x - 2} \right)}}.\dfrac{{x\left( {x - 2} \right)}}{{ - 4}}\\
= - \dfrac{{3x - 4}}{4}\\
\left| B \right| = 2x + 5\\
\to \left[ \begin{array}{l}
- \dfrac{{3x - 4}}{4} = 2x + 5\left( {DK: \dfrac{4}{3} \ge x} \right)\\
\dfrac{{3x - 4}}{4} = 2x + 5\left( {DK:x > \dfrac{4}{3}} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
- 3x + 4 = 8x + 20\\
3x - 4 = 8x + 20
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{{16}}{{11}}\\
x = - \dfrac{{24}}{5}\left( l \right)
\end{array} \right.\\
b)\left( {2 - x} \right).B = \left( {2 - x} \right).\left( { - \dfrac{{3x - 4}}{4}} \right)\\
= - \dfrac{{ - 3{x^2} + 10x - 8}}{4}\\
= \dfrac{{3{x^2} - 10x + 8}}{4}\\
= \dfrac{{3{x^2} - 2.x\sqrt 3 .\dfrac{5}{{\sqrt 3 }} + \dfrac{{25}}{3} - \dfrac{1}{3}}}{4}\\
= \dfrac{{{{\left( {x\sqrt 3 - \dfrac{5}{{\sqrt 3 }}} \right)}^2} - \dfrac{1}{3}}}{4}\\
Do:{\left( {x\sqrt 3 - \dfrac{5}{{\sqrt 3 }}} \right)^2} \ge 0\forall x\\
\to {\left( {x\sqrt 3 - \dfrac{5}{{\sqrt 3 }}} \right)^2} - \dfrac{1}{3} \ge - \dfrac{1}{3}\\
\to \dfrac{{{{\left( {x\sqrt 3 - \dfrac{5}{{\sqrt 3 }}} \right)}^2} - \dfrac{1}{3}}}{4} \ge - \dfrac{1}{{12}}\\
\to Min = - \dfrac{1}{{12}}\\
\Leftrightarrow x\sqrt 3 - \dfrac{5}{{\sqrt 3 }} = 0\\
\to x = \dfrac{5}{3}
\end{array}\)
c) Để B có giá trị nguyên âm lớn nhất
⇔ \({3x - 4}\) ∈ B(4) và mang giá trị lớn nhất
⇒ x thuộc giá trị dương vô cùng
⇒ Không tồn tại x TMĐK
\(\begin{array}{l}
d)\left| { - \dfrac{{3x - 4}}{4}} \right| + 3 < 2x - 1\\
\to \left| { - \dfrac{{3x - 4}}{4}} \right| < 2x - 4\\
\to \left[ \begin{array}{l}
- \dfrac{{3x - 4}}{4} < 2x - 4\left( {DK:\dfrac{4}{3} \ge x} \right)\\
\dfrac{{3x - 4}}{4} > 2x - 4\left( {DK:\dfrac{4}{3} < x} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
- 3x + 4 < 8x - 16\\
3x - 4 > 8x - 16
\end{array} \right.\\
\to \left[ \begin{array}{l}
\dfrac{{20}}{{11}} < x\left( l \right)\\
\dfrac{{12}}{5} > x
\end{array} \right.\\
\to \dfrac{{12}}{5} > x > \dfrac{4}{3}
\end{array}\)
