Tính giới hạn \[I = \mathop {\lim }\limits_{x \to 0} \frac{{\arctan \left( {{x^2} + 4x} \right) + \ln \left( {1 + 3\tan x} \right) - {x^2}}}{{\arctan \left( {4x} \right) + \cos 2x - {e^x}}}\]

A. 𝑡 = 1000.

B. 𝑡 = 1750.

C. 𝑡 = 375.

D. 𝑡 = 500.

A. \[\frac{1}{{2\pi }}\left( {0,16} \right){m^3}/s\]

B. \[\frac{1}{{2\pi }}\left( {0,16} \right)m/s\]

C. \[\frac{1}{{2\pi }}\left( {0,08} \right)m/s\]

D. \[\frac{1}{{2\pi }}\left( {0,08} \right){m^3}/s\]

A. \[\frac{{20}}{\pi }cm\]

B. \[\frac{{10}}{\pi }cm\]

C. \[\frac{{30}}{{{\pi ^2}}}cm\]

D. 20 cm

A. I = 2

B. I = 1

C. \[I = \frac{1}{2}\]

D. I = 0

A. Q = 50

B. Q = 49

C. Q = 52

D. cả A, B, C đều sai

A. \[f\left( x \right) = \frac{1}{2}x + \frac{3}{8}{x^2} - \frac{1}{{48}}{x^3} + o\left( {{x^3}} \right)\]

B. \[f\left( x \right) = \frac{1}{2}x + \frac{3}{8}{x^2} + \frac{1}{{48}}{x^3} + o\left( {{x^3}} \right)\]

C. \[f\left( x \right) = \frac{1}{2}x - \frac{3}{8}{x^2} - \frac{1}{{48}}{x^3} + o\left( {{x^3}} \right)\]

D. \[f\left( x \right) = \frac{1}{2}x + \frac{1}{8}{x^2} - \frac{1}{{48}}{x^3} + o\left( {{x^3}} \right)\]

A. \[f\left( x \right) = 1 - \frac{1}{3}\left( {x - 1} \right) - \frac{2}{9}{\left( {x - 1} \right)^2} + o\left( {{{\left( {x - 1} \right)}^2}} \right)\]

B. \[f\left( x \right) = 1 + \frac{1}{3}\left( {x - 1} \right) - \frac{2}{9}{\left( {x - 1} \right)^2} + o\left( {{{\left( {x - 1} \right)}^2}} \right)\]

C. \[f\left( x \right) = 1 - \frac{1}{3}\left( {x - 1} \right) + \frac{2}{9}{\left( {x - 1} \right)^2} + o\left( {{{\left( {x - 1} \right)}^2}} \right)\]

D. \[f\left( x \right) = 1 - \frac{1}{3}\left( {x - 1} \right) + \frac{4}{9}{\left( {x - 1} \right)^2} + o\left( {{{\left( {x - 1} \right)}^2}} \right)\]