Câu hỏi trắc nghiệm (10 câu):
-
- A. \(x > - 2\) và \(x \ne - 1\)
- B. \(x > - 2\) và \(x < \frac{4}{3}\)
- C. \(x > - 2,x \ne - 1\) và \(x \le \frac{4}{3}\)
- D. \(x \ne - 2\) và \(x \ne - 1\)
-
- A. \(\left\{ { - \frac{2}{m}} \right\}\)
- B. \(\emptyset \)
- C. R
- D. \(R\backslash \left\{ 0 \right\}\)
-
- A. \(\left( { - \frac{{39}}{{26}};\frac{3}{{13}}} \right)\)
- B. \(\left( { - \frac{{17}}{{13}}; - \frac{5}{{13}}} \right)\)
- C. \(\left( {\frac{{39}}{{26}};\frac{1}{2}} \right)\)
- D. \(\left( { - \frac{1}{3};\frac{{17}}{6}} \right)\)
-
- A. \(\left( { - 10;7;9} \right)\)
- B. \(\left( {\frac{3}{2}; - 2;\frac{3}{2}} \right)\)
- C. \(\left( { - \frac{1}{4}; - \frac{9}{4};\frac{5}{4}} \right)\)
- D. \(\left( { - 5; - 7; - 8} \right)\)
-
- A. \(3{x^2} + 5 = - 2\sqrt {x - 1} \)
- B. \({x^2} - 3\sqrt {1 - x} = 4 + \sqrt {x - 5} \)
- C. \({x^2} + 2 = \sqrt {x + 4} \)
- D. \({x^2} + 4x + 6 = 0\)
-
- A. Ø
- B. {Ø}
- C. {13/9}
- D. {13/9;14/9}
-
- A. \({x^2} + x + 1 = 0\)
- B. \({x^2} - 4x + 4 = 0\)
- C. \({x^5} + 3{x^2} + x - 5 = 0\)
- D. \(\sqrt {x - 2} + x = 1 + \sqrt {x - 2} \)
-
- A. {Ø}
- B. \(\left\{ {\frac{3}{2}} \right\}\)
- C. \(\left\{ {-\frac{3}{2}} \right\}\)
- D. \(( - \infty ; + \infty )\)
-
- A. (2; –2), (3; –3)
- B. (–2; 2), (–3; 3)
- C. (1; –1), (3; –3)
- D. (–1; 1), (–4; 4)
-
- A. \(\left( {1;2} \right),\left( {\sqrt 2 ;\sqrt 2 } \right)\)
- B. \(\left( {2;1} \right),\left( {\sqrt 3 ;\sqrt 3 } \right)\)
- C. \(\left( {\frac{2}{3};3} \right),\left( {\sqrt 3 ;\frac{2}{{\sqrt 3 }}} \right)\)
- D. \(\left( {\frac{1}{2};1} \right),\left( {\frac{{\sqrt 2 }}{3};\sqrt 3 } \right)\)
VIDEO
YOMEDIA
ADMICRO
NONE
