Đáp án:
$\begin{array}{l}
1)a)\dfrac{{2x - 1}}{3} - x = \dfrac{{x + 2}}{4}\\
\Rightarrow \dfrac{{4.\left( {2x - 1} \right) - 12x}}{{12}} = \dfrac{{3\left( {x + 2} \right)}}{{12}}\\
\Rightarrow 8x - 4 - 12x = 3x + 6\\
\Rightarrow 7x = - 10\\
\Rightarrow x = \dfrac{{ - 10}}{7}\\
b)Dkxd:x \ne - 1;x \ne 2\\
\dfrac{1}{{x + 1}} - \dfrac{5}{{x - 2}} = \dfrac{{15}}{{\left( {x + 1} \right)\left( {2 - x} \right)}}\\
\Rightarrow \dfrac{{x - 2 - 5\left( {x + 1} \right)}}{{\left( {x + 1} \right)\left( {x - 2} \right)}} = \dfrac{{ - 15}}{{\left( {x + 1} \right)\left( {x - 2} \right)}}\\
\Rightarrow x - 2 - 5x - 5 = - 15\\
\Rightarrow 4x = 8\\
\Rightarrow x = 2\left( {ktm} \right)\\
\Rightarrow x \in \emptyset \\
c)\dfrac{1}{9}{\left( {x - 3} \right)^2} - \dfrac{1}{{25}}{\left( {x + 5} \right)^2} = 0\\
\Rightarrow {\left( {\dfrac{{x - 3}}{3}} \right)^2} = {\left( {\dfrac{{x + 5}}{5}} \right)^2}\\
\Rightarrow \left[ \begin{array}{l}
\dfrac{{x - 3}}{3} = \dfrac{{x + 5}}{5}\\
\dfrac{{x - 3}}{3} = - \dfrac{{x + 5}}{5}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
5x - 15 = 3x + 15\\
5x - 15 = - 3x - 15
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
2x = 30\\
8x = 0
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 15\\
x = 0
\end{array} \right.\\
2)a)Dkxd:x \ne - 3;x \ne 3;x \ne - 1\\
P = \left( {\dfrac{{2x}}{{x + 3}} + \dfrac{x}{{x - 3}} - \dfrac{{3{x^2} + 3}}{{{x^2} - 9}}} \right):\left( {\dfrac{{2x - 2}}{{x - 3}} - 1} \right)\\
= \dfrac{{2x\left( {x - 3} \right) + x\left( {x + 3} \right) - 3{x^2} - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}:\dfrac{{2x - 2 - x + 3}}{{x - 3}}\\
= \dfrac{{2{x^2} - 6x + {x^2} + 3x - 3{x^2} - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\dfrac{{x - 3}}{{x + 1}}\\
= \dfrac{{ - 3x - 3}}{{x + 3}}.\dfrac{1}{{x + 1}}\\
= \dfrac{{ - 3}}{{x + 3}}\\
b)x = - \dfrac{1}{3}\left( {tmdk} \right)\\
P = \dfrac{{ - 3}}{{\dfrac{{ - 1}}{3} + 3}} = \dfrac{{ - 3}}{{\dfrac{8}{3}}} = \dfrac{{ - 9}}{8}\\
c)P.\dfrac{1}{x} = \dfrac{{ - 3}}{{x + 3}}.\dfrac{1}{x} = \dfrac{{ - 3}}{{{x^2} + 3x}}\\
Do:{x^2} + 3x\\
= {x^2} + 2.x.\dfrac{3}{2} + \dfrac{9}{4} - \dfrac{9}{4}\\
= {\left( {x + \dfrac{3}{2}} \right)^2} - \dfrac{9}{4} \ge - \dfrac{9}{4}\\
\Rightarrow \dfrac{1}{{{x^2} + 3x}} \le \dfrac{{ - 4}}{9}\\
\Rightarrow \dfrac{{ - 3}}{{{x^2} + 3x}} \ge \dfrac{{12}}{9} = \dfrac{4}{3}\\
\Rightarrow P.\dfrac{1}{x} \ge \dfrac{4}{3}\\
\Rightarrow GTNN = \dfrac{4}{3}\,khi:x = - \dfrac{3}{2}\left( {tmdk} \right)\\
d)P < \dfrac{1}{2}\\
\Rightarrow \dfrac{{ - 3}}{{x + 3}} < \dfrac{1}{2}\\
\Rightarrow \dfrac{{ - 3}}{{x + 3}} - \dfrac{1}{2} < 0\\
\Rightarrow \dfrac{{ - 6 - x - 3}}{{2\left( {x + 3} \right)}} < 0
\end{array}$$\begin{array}{l}
\Rightarrow \dfrac{{x + 9}}{{2\left( {x + 3} \right)}} > 0\\
\Rightarrow \left[ \begin{array}{l}
x > - 3\\
x < - 9
\end{array} \right.
\end{array}$
B3) Đổi 20 phút = 1/3 giờ
Gọi quãng đường lúc đi là: x (km) (x>0)
=> thời gian lúc đi là : $\dfrac{x}{9}\left( h \right)$
=> quãng đường lúc về là: x+6 (km)
=> thời gian về là: $\dfrac{{x + 6}}{{12}}\left( h \right)$
Ta có pt thời gian:
$\begin{array}{l}
\dfrac{x}{9} - \dfrac{{x + 6}}{{12}} = \dfrac{1}{3}\\
\Rightarrow \dfrac{x}{9} - \dfrac{x}{{12}} - \dfrac{1}{2} = \dfrac{1}{3}\\
\Rightarrow \dfrac{x}{{36}} = \dfrac{5}{6}\\
\Rightarrow x = 30\left( {km} \right)\left( {tmdk} \right)
\end{array}$
Vậy lúc đi quãng đường là: 30 km.
