Đề kiểm tra 1 tiết Chương 2 Giải tích 12 năm 2018...

A. \({a^{\frac{7}{{10}}}}\)

B. \({a^{ - \frac{8}{5}}}\)

C. \({a^{\frac{1}{6}}}\)

D. \({a^{\frac{{ - 16}}{5}}}\)

A. a + 5

B. a - 4

C. 3 + a

D. 2a + 3

A. \(y' = {7^{2{x^3} + 3x - 4}}\ln 7\)

B. \(y' = (6{x^2} + 3){7^{2{x^3} + 3x - 4}}\ln 7\)

C. \(y' = (6{x^2} + 3){7^{2{x^3} + 3x - 5}}\)

D. \(y' = (2{x^3} + 3x - 4){7^{2{x^3} + 3x - 5}}\)

A. \(y=5x + 5\ln 2x\)

B. \(y = 5x\ln (2x)\)

C. \(y = 5 + \frac{5}{{2x}}\)

D. \(y=5x + {\ln ^2}2x\)

A. \(y' = \frac{5}{{(5x + 3)\ln 4}}\)

B. \(y' = \frac{5}{{5x + 3}}\)

C. \(y' = \frac{1}{{(5x + 3)\ln 4}}\)

D. \(y' = \frac{1}{{\ln 4}}\)

A. \({P_{\min }} = \frac{{2\sqrt {11}  - 3}}{3} \cdot \)

B. \({P_{\min }} = \frac{{9\sqrt {11}  - 19}}{9} \cdot \)

C. \({P_{\min }} = \frac{{9\sqrt {11}  + 19}}{9} \cdot \)

D. \({P_{\min }} = \frac{{18\sqrt {11}  - 29}}{{21}} \cdot \)

A. \(\log (a + b) = 2 + \log a + \log b.\)

B. \(\log a + {\mathop{\rm logb}\nolimits}  = 2.\)

C. \(\log (a + b) = \frac{1}{2}\left( {1 + \log a + \log b} \right).\)

D. \(log(a + b) = 1 + \frac{1}{2}\left( {\log a + \log b} \right).\)

A. \(\frac{1}{t} \le t\)

B. \(0 < \frac{1}{t} \le t\)

C. \(0 < t \le \frac{1}{t}\)

D. \(0 < t \le {t^2}\)

A. (5;10)

B. \(\left[ {0;6} \right).\)

C. \(\left( { - \infty ;0} \right).\)

D. \(\left[ {10; + \infty } \right).\)

A. \(S = ( - 1;1)\)

B. \(S = \left( { - \infty ; - 1} \right) \cup \left( {1;3} \right)\)

C. \(S = ( - \infty ; - 3) \cup (1; + \infty )\)

D. \(S = ( - \infty ; - 2) \cup (0;2)\)

A. m = 10

B. m = - 5

C. m = 32

D. m = 5