Câu hỏi:

\({{x}^{2}}+1+C\)

\({{x}^{2}}+x+C\)

\({{x}^{2}}+C\)

\(2x+C\)

\(\int{f\left( x \right)\text{d}x=F\left( x \right)}+C\)

\({{\left( \int{f\left( x \right)\text{d}x} \right)}^{\prime }}=f\left( x \right)\)

\({{\left( \int{f\left( x \right)\text{d}x} \right)}^{\prime }}=f\left( x \right)+C\)

\({{\left( \int{f\left( x \right)\text{d}x} \right)}^{\prime }}={F}'\left( x \right)\)

5

3

-3

4

\(S=\int\limits_{0}^{1}{\left( {{x}^{2}}+1 \right)\text{d}x}\)

\(S=\pi \int\limits_{0}^{1}{\left( {{x}^{2}}+1 \right)\text{d}x}\)

\(S=\int\limits_{0}^{1}{{{\left( {{x}^{2}}+1 \right)}^{2}}\text{d}x}\)

\(S=\pi \int\limits_{0}^{1}{\left| {{x}^{2}}-1 \right|\text{d}x}\)

\(\overrightarrow{{{n}_{1}}}=\left( -1;1;1 \right)\)

\(\overrightarrow{{{n}_{2}}}=\left( 1;-1;1 \right)\)

\(\overrightarrow{{{n}_{3}}}=\left( 1;1;1 \right)\)

\(\overrightarrow{{{n}_{4}}}=\left( 1;1;-1 \right)\)

\(2x+3y+z+6=0\)

\(x-3y+z+6=0\)

\(x-3y+z-6=0\)

\(-x+3y-z+5=0\)

\(\left\{ \begin{matrix} x=1+t \\ y=1+2t \\ z=1-3t \\ \end{matrix}\left( t\in \mathbb{R} \right) \right.\)

\(\left\{ \begin{matrix} x=1+3t \\ y=1+2t \\ z=1+t \\ \end{matrix}\left( t\in \mathbb{R} \right) \right.\)

\(\left\{ \begin{matrix} x=1+t \\ y=1+3t \\ z=1+2t \\ \end{matrix}\left( t\in \mathbb{R} \right) \right.\)

\(\left\{ \begin{matrix} x=1+t \\ y=1+2t \\ z=1+3t \\ \end{matrix}\left( t\in \mathbb{R} \right) \right.\)

\(\overrightarrow{{{u}_{1}}}=\left( -1;2;0 \right)\)

\(\overrightarrow{{{u}_{2}}}=\left( 1;3;1 \right)\)

\(\overrightarrow{{{u}_{3}}}=\left( 1;2;1 \right)\)

\(\overrightarrow{{{u}_{4}}}=\left( -1;2;1 \right)\)

\(\cos \alpha =\frac{1}{3}\)

\(\cos \alpha =-\frac{1}{3}\)

\(\cos \alpha =-\frac{11}{15}\)

\(\cos \alpha =\frac{11}{15}\)

\(I\left( 1;4;-2 \right),R=3\)

\(I\left( -1;-4;2 \right),R=3\)

\(I\left( 1;4;-2 \right),R=9\)

\(I\left( -1;-4;2 \right),R=9\)