Trong các động vật sau, con nào tại có màng nhĩ ?
A.Cá mực.
B.Cá chép.
C.Thằn lằn.
D.Ếch nhái.

Trong các đặc điểm sau của thằn lằn, đặc điểm nào giúp thằn lằn mang được các giác quan nằm trên đầu, tạo điều kiện bắt mồi
A.Da khô, có vảy sừng bao bọc,
B.Có cổ dài.
C.Mắt có mi cử động, có nước mắt,
D.Màng nhĩ nằm trong một hốc nhỏ bên đầu

\(\mathop {\lim }\limits_{x \to \pm \infty } \left( {3x + 2 - \dfrac{{{x^2} - x + 1}}{{x + 2}}} \right)\)
A.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{x\left( {2 + \dfrac{9}{x} + \dfrac{3}{{{x^2}}}} \right)}}{{1 + \dfrac{2}{x}}} = \dfrac{3}{2} \).
B.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{x\left( {2 + \dfrac{9}{x} + \dfrac{3}{{{x^2}}}} \right)}}{{1 + \dfrac{2}{x}}} = \pm \infty \).
C.\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{x\left( {2 + \dfrac{9}{x} + \dfrac{3}{{{x^2}}}} \right)}}{{1 + \dfrac{2}{x}}} = + \infty \). \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{x\left( {2 + \dfrac{9}{x} + \dfrac{3}{{{x^2}}}} \right)}}{{1 + \dfrac{2}{x}}} = - \dfrac{3}{2} \).
D.\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{x\left( {2 + \dfrac{9}{x} + \dfrac{3}{{{x^2}}}} \right)}}{{1 + \dfrac{2}{x}}} = + \dfrac{3}{2} \). \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{x\left( {2 + \dfrac{9}{x} + \dfrac{3}{{{x^2}}}} \right)}}{{1 + \dfrac{2}{x}}} = - \dfrac{3}{2} \).

\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{\left( {x - 1} \right)\left( {x + 3} \right)}}{{1 - 3\left| x \right|}}\)
A.\( \mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{\left( {x - 1} \right)\left( {x + 3} \right)}}{{1 - 3\left| x \right|}} = - \infty \).
B.\( \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\left( {x - 1} \right)\left( {x + 3} \right)}}{{1 - 3\left| x \right|}} = + \infty \). \( \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {x - 1} \right)\left( {x + 3} \right)}}{{1 - 3\left| x \right|}} = - \infty \).
C.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{\left( {x - 1} \right)\left( {x + 3} \right)}}{{1 - 3\left| x \right|}} = \dfrac{1}{3} \).
D.\( \mathop {\lim }\limits_{x \to + \infty } \dfrac{{\left( {x - 1} \right)\left( {x + 3} \right)}}{{1 - 3\left| x \right|}} = - \dfrac{2}{3} \). \( \mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {x - 1} \right)\left( {x + 3} \right)}}{{1 - 3\left| x \right|}} = \dfrac{1}{3} \).

\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{{x^9} + 6x + 8}}{{4\left| {{x^9}} \right| - 2x + 3}}\)
A.\(\dfrac{1}{4}\).
B.\(- \dfrac{1}{4}\).
C.\(0.\)
D.\(+ \infty\)

\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{\left( {3x - 1} \right)\sqrt {{x^6} + x + 1} }}{{\sqrt {{x^8} - x + 2} }}\)
A.\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{\left( {3x - 1} \right)\sqrt {{x^6} + x + 1} }}{{\sqrt {{x^8} - x + 2} }} = \dfrac{2}{3}\).
B.\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{\left( {3x - 1} \right)\sqrt {{x^6} + x + 1} }}{{\sqrt {{x^8} - x + 2} }} = \pm\infty\).
C.\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{\left( {3x - 1} \right)\sqrt {{x^6} + x + 1} }}{{\sqrt {{x^8} - x + 2} }} = 3\), \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {3x - 1} \right)\sqrt {{x^6} + x + 1} }}{{\sqrt {{x^8} - x + 2} }} = - 3\).
D.\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{\left( {3x - 1} \right)\sqrt {{x^6} + x + 1} }}{{\sqrt {{x^8} - x + 2} }} = \dfrac{2}{3}\), \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{\left( {3x - 1} \right)\sqrt {{x^6} + x + 1} }}{{\sqrt {{x^8} - x + 2} }} = - \infty\).

\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}}\)
A.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = \pm \infty \).
B.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = \pm 1\) .
C.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = - \infty \).
D.\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} \) không tồn tại.
\(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = +\infty \).

\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2x\sqrt {{x^2} + x} - \sqrt {9{x^2} + 1} }}{{\left( {2x + 1} \right)\left( {x - 2} \right)}}\)
A.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2x\sqrt {{x^2} + x} - \sqrt {9{x^2} + 1} }}{{\left( {2x + 1} \right)\left( {x - 2} \right)}} = \dfrac{1}{2}\)
B.\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{2x\sqrt {{x^2} + x} - \sqrt {9{x^2} + 1} }}{{\left( {2x + 1} \right)\left( {x - 2} \right)}} = \dfrac{1}{2}\), \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{2x\sqrt {{x^2} + x} - \sqrt {9{x^2} + 1} }}{{\left( {2x + 1} \right)\left( {x - 2} \right)}} = - \infty\).
C.\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{2x\sqrt {{x^2} + x} - \sqrt {9{x^2} + 1} }}{{\left( {2x + 1} \right)\left( {x - 2} \right)}} = \dfrac{1}{2}\), \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{2x\sqrt {{x^2} + x} - \sqrt {9{x^2} + 1} }}{{\left( {2x + 1} \right)\left( {x - 2} \right)}} = - 1\).
D.\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{2x\sqrt {{x^2} + x} - \sqrt {9{x^2} + 1} }}{{\left( {2x + 1} \right)\left( {x - 2} \right)}} = 1\), \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{2x\sqrt {{x^2} + x} - \sqrt {9{x^2} + 1} }}{{\left( {2x + 1} \right)\left( {x - 2} \right)}} = - 1\).

\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}}\)
A.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = \dfrac{1}{9}\) .
B.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = + \infty \).
C.\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}}\) không tồn tại. \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = \dfrac{1}{9}\) .
D.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = \mp \infty \).

\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}}\)
A.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = \dfrac{1}{9}\) .
B.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = + \infty \).
C.\(\mathop {\lim }\limits_{x \to + \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}}\) không tồn tại. \(\mathop {\lim }\limits_{x \to - \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = \dfrac{1}{9}\) .
D.\(\mathop {\lim }\limits_{x \to \pm \infty } \dfrac{{2{x^2} + 3x - 5}}{{\sqrt {{x^2} + 1} - 3x}} = \mp \infty \).