Câu 18: Đáp án C
Kẻ $AP\bot \Delta \Rightarrow P\left( t+2;1-2t;2t \right)\Rightarrow \overrightarrow{AP}=\left( t+3;-2t;2t-6 \right)$
Ta có $\overrightarrow{{{u}_{\Delta }}}=\left( 1;-2;2 \right),AP\bot \Delta \Leftrightarrow \overrightarrow{AP}.\overrightarrow{{{u}_{\Delta }}}=0\Leftrightarrow \left( t+3 \right)+4t+2\left( 2t-6 \right)=0\Leftrightarrow t=1\Rightarrow P\left( 3;-1;2 \right)$
Câu 19: Đáp án D
Kẻ $BP\bot AC\Rightarrow BP\bot \left( SAC \right)\Rightarrow \widehat{\left( SB;\left( SAC \right) \right)}=\widehat{BSP}$
$BP=\dfrac{AB.BC}{AC}=\dfrac{aa\sqrt{3}}{2a}=\dfrac{a\sqrt{3}}{2}$
$SB=\sqrt{S{{A}^{2}}+A{{B}^{2}}}=a\sqrt{3}$
$\Rightarrow \sin \widehat{BSP}=\dfrac{BP}{SB}=\dfrac{1}{2}\Rightarrow \widehat{BSP}=30{}^\circ $
Câu 20: Đáp án A
Ta có ${y}'=\dfrac{1}{3}{{\left( {{x}^{2}}+x+1 \right)}^{\dfrac{1}{3}-1}}.\left( 2x+1 \right)=\dfrac{2x+1}{3\sqrt[3]{{{\left( {{x}^{2}}+x+1 \right)}^{2}}}}$
Câu 21: Đáp án B
Kẻ $SH\bot AB\Rightarrow SH\bot \left( ABCD \right)$.
Ta có $AD//BC\Rightarrow AD\bot \left( SBC \right)$
$\Rightarrow d\left( AD,SC \right)=d\left( A;\left( SBC \right) \right)=2d\left( H;\left( SBC \right) \right)=2HP.$
Trong đó $HP\bot \text{S}B.$
Cạnh $SH=\sqrt{S{{A}^{2}}-A{{H}^{2}}}=\sqrt{S{{A}^{2}}-{{\left( \dfrac{AB}{2} \right)}^{2}}}=2a$
$\Rightarrow HP=\dfrac{HS.HB}{SB}=\dfrac{2a.a}{a\sqrt{5}}\Rightarrow d\left( AD;SC \right)=\dfrac{4a}{\sqrt{5}}$.
Câu 22: Đáp án B
Ta có $I=\dfrac{1}{2}.\left. \dfrac{{{3}^{2\text{x}+1}}}{\ln 3} \right|_{0}^{1}=\dfrac{12}{\ln 3}$.
Câu 23: Đáp án C
Ta có $y' = 2\left( {{x^2} - x} \right)\left( {2{\rm{x}} - 1} \right) < 0 \Leftrightarrow \left[ {\begin{array}{*{20}{l}}
{x < 0}\\
{\frac{1}{2} < x < 1}
\end{array}} \right.$
Câu 24: Đáp án A
Ta có $y = x + \frac{4}{{x + 1}} \Rightarrow y' = 1 - \frac{4}{{{{\left( {x + 1} \right)}^2}}};\,\left\{ \begin{array}{l}
x \in \left( {0;2} \right)\\
y' = 0
\end{array} \right. \Leftrightarrow x = 1$
Tính $y\left( 0 \right) = 4;\,y\left( 2 \right) = \frac{{10}}{3};\,y\left( 1 \right) = 3 \Rightarrow \left\{ \begin{array}{l}
a = 3\\
A = 4
\end{array} \right. \Rightarrow a + A = 7.$
Câu 25: Đáp án A
Ta có $\left\{ {\begin{array}{*{20}{l}}
{{z_1} + {z_2} = 6}\\
{{z_1}{z_2} = 13}
\end{array}} \right. \Rightarrow {z^2} - 6z + 13 = 0.$
Câu 26: Đáp án A
Ta có $F\left( x \right)=-\int{\ln \left( x+3 \right)d\left( \dfrac{1}{x} \right)=-\dfrac{\ln \left( x+3 \right)}{x}+\int{\dfrac{1}{x}d\left[ \ln \left( x+3 \right) \right]}}$
$=-\dfrac{\ln \left( x+3 \right)}{x}+\int{\dfrac{1}{x}.\dfrac{1}{x+3}d\text{x}=-\dfrac{\ln \left( x+3 \right)}{x}+\dfrac{1}{3}\int{\left( \dfrac{1}{x}-\dfrac{1}{x+3} \right)d\text{x}}}$
$=-\dfrac{\ln \left( x+3 \right)}{x}+\dfrac{1}{3}\ln \left| \dfrac{x}{x+3} \right|+C.$
Mà $F\left( -2 \right)+F\left( 1 \right)=0\Rightarrow \left( \dfrac{1}{3}\ln 2+{{C}_{1}} \right)+\left( -\ln 4+\dfrac{1}{3}\ln \dfrac{1}{4}+{{C}_{2}} \right)=0\Rightarrow -\dfrac{7}{3}\ln 2+{{C}_{1}}+{{C}_{2}}=0$
$\Rightarrow F\left( -1 \right)+F\left( 2 \right)=\left( \ln 2+\dfrac{1}{3}\ln \dfrac{1}{2}+{{C}_{1}} \right)+\left( -\dfrac{1}{2}\ln 5+\dfrac{1}{3}\ln \dfrac{2}{5}+{{C}_{2}} \right)=\dfrac{10}{3}\ln 2-\dfrac{5}{6}\ln 5.$
