Câu hỏi:

\(y = {x^2}\)

\(y = - {x^2}\)

\(y = 2x\)

\(y = - 2x\)

\(x = \frac{1}{6}\)

\(x = - \frac{1}{6}\)

\(x = \frac{1}{3}\)

\(x = - \frac{1}{3}\)

\(f\left( x \right) \le 0,\forall x \in \mathbb{R}\)

\(f\left( x \right) < 0,\forall x \in \mathbb{R}\)

\(f\left( x \right) > 0,\forall x \in \mathbb{R}\)

\(f\left( x \right) \ge 0,\forall x \in \mathbb{R}\)

\(x = 4\)

\(x = 2\)

\(x = 1\)

\(x = 3\)

\(\overrightarrow {{n_1}} = \left( { - 5;7} \right)\)

\(\overrightarrow {{n_2}} = \left( {5;3} \right)\)

\(\overrightarrow {{n_3}} = \left( {3; - 5} \right)\)

\(\overrightarrow {{n_4}} = \left( {3;7} \right)\)

Trùng nhau

Cắt nhau nhưng không vuông góc

Vuông góc với nhau

Song song

\(I\left( { - 3;4} \right),R = 7\)

\(I\left( {3; - 4} \right),R = 7\)

\(I\left( {3; - 4} \right),R = 49\)

\(I\left( { - 3;4} \right),R = 49\)

\(\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1\)

\(\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1\)

\(\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 0\)

\(\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 0\)

4 cách

6 cách

7 cách

12 cách

\(A_n^k = \frac{{n!}}{{k!\left( {n + k} \right)!}}\)

\(A_n^k = \frac{{n!}}{{\left( {n + k} \right)!}}\)

\(A_n^k = \frac{{n!}}{{k!\left( {n - k} \right)!}}\)

\(A_n^k = \frac{{n!}}{{\left( {n - k} \right)!}}\)