Trong không gian với hệ tọa độ Oxyz, cho ba vector $\vec a = \left( {2;3; - 5} \right);{\mkern 1mu} {\mkern 1mu} \vec b = \left( {0; - 3;4} \right);{\mkern 1mu} {\mkern 1mu} \vec c = \left( {1; - 2;3} \right)$. Tọa độ vector $\vec n = 3\vec a + 2\vec b - \vec c$ là:

\(m > 6\)

\(m \ge 6\)

\(m \le 6\)

\(m < 6\)

\({(x + 1)^2} + {(y + 2)^2} + {(z - 3)^2} = 53.\)

\({(x + 1)^2} + {(y + 2)^2} + {(z + 3)^2} = 53.\)

\({(x - 1)^2} + {(y - 2)^2} + {(z - 3)^2} = 53.\)

\({(x - 1)^2} + {(y - 2)^2} + {(z + 3)^2} = 53.\)

$I = \left. {\dfrac{{{x^2}\ln \left( {x + 2} \right)}}{2}} \right|_0^1 - \dfrac{1}{2}\int\limits_0^1 {\dfrac{{{x^2}}}{{x + 2}}{\rm{d}}x} .$

$I = \left. {{x^2}\ln \left( {x + 2} \right)} \right|_0^1 - \dfrac{1}{4}\int\limits_0^1 {\dfrac{{{x^2}}}{{x + 2}}{\rm{d}}x} .$

$I = \left. {\dfrac{{{x^2}\ln \left( {x + 2} \right)}}{2}} \right|_0^1 + \int\limits_0^1 {\dfrac{{{x^2}}}{{x + 2}}{\rm{d}}x} .$

$I = \left. {\dfrac{{{x^2}\ln \left( {x + 2} \right)}}{4}} \right|_0^1 - \dfrac{1}{4}\int\limits_0^1 {\dfrac{{{x^2}}}{{x + 2}}{\rm{d}}x} .$

\(\int\limits_a^b {udv}  = \left. {uv} \right|_a^b - \int\limits_a^b {vdu} \)

\(\int\limits_a^b {udv}  = \left. {uv} \right|_a^b + \int\limits_a^b {vdu} \)

\(\int\limits_a^b {udv}  = \left. {uv} \right|_b^a - \int\limits_a^b {vdu} \)

\(\int\limits_a^b {udv}  = \left. {uv} \right|_a^b - \int\limits_b^a {vdu} \)

\(\left( {1;1; - 1} \right)\)

\(\left( { - 1;1; - 1} \right)\)    

\(\left( {1;1; - \dfrac{1}{3}} \right)\)

\(\left( {3;3; - 1} \right)\)

$I = {x^2}\sin x|_0^\pi  - \int\limits_0^\pi  {x\sin xdx} $

$I = {x^2}\sin x|_0^\pi  + 2\int\limits_0^\pi  {x\sin xdx} $

$I = {x^2}\sin x|_0^\pi  + 2\int\limits_0^\pi  {x\sin xdx} $

$I = {x^2}\sin x|_0^\pi  - 2\int\limits_0^\pi  {x\sin xdx} $