Vì sao tôm cần phải lột xác để lớn?
A.Lớp vỏ kitin cứng ngăn tôm lớn lên
B.Lớp vỏ kitin cũ xấu
C.Lớp vỏ kitin cũ dễ vỡ
D.Tôm lột xác không vì lý do nào cả.

\(y = \frac{{x + 2}}{{{x^2} - 4x + 3}}\)
A.\(D = \mathbb{R}\backslash \left\{ {1;\,\,3} \right\}.\)
B.\(D = \mathbb{R}\backslash \left\{ { - 1;\,\,3} \right\}.\)
C.\(D = \mathbb{R}\backslash \left\{ {1;\,\, - 3} \right\}.\)
D.\(D = \mathbb{R}\backslash \left\{ { - 1;\,\, - 3} \right\}.\)

\(y = \frac{{x + 2}}{{{x^2} - 4x + 3}}\)
A.\(D = \mathbb{R}\backslash \left\{ {1;\,\,3} \right\}.\)
B.\(D = \mathbb{R}\backslash \left\{ { - 1;\,\,3} \right\}.\)
C.\(D = \mathbb{R}\backslash \left\{ {1;\,\, - 3} \right\}.\)
D.\(D = \mathbb{R}\backslash \left\{ { - 1;\,\, - 3} \right\}.\)

\(y = \frac{{\sqrt {x + 2} }}{{\left| x \right| - 1}}\)
A.\(D = \left[ { - 2; + \infty } \right)\backslash \left\{ 1 \right\}.\)
B.\(D = \left[ { - 2; + \infty } \right)\backslash \left\{ { - 1;\,\,1} \right\}.\)
C.\(D = \left( { - 2; + \infty } \right)\backslash \left\{ { - 1;\,\,1} \right\}.\)
D.\(D = \left( { - 2; + \infty } \right)\backslash \left\{ 1 \right\}.\)

\(y = \sqrt {6 - x} + \frac{1}{{{x^2} - 4}}\)
A.\(D = \left[ {6; + \infty } \right)\backslash \left\{ { - 2;\,\,2} \right\}.\)
B.\(D = \left( { - \infty ;6} \right)\backslash \left\{ { - 2;\,\,2} \right\}.\)
C.\(D = \left( {6; + \infty } \right)\backslash \left\{ { - 2;\,\,2} \right\}.\)
D.\(D = \left( { - \infty ;\,\,6} \right]\backslash \left\{ { - 2;\,\,2} \right\}.\)

\(y = \frac{{3x + 5}}{{\left| {x + 1} \right| + \left| {x - 1} \right|}}\)
A.\(D = \mathbb{R}\backslash \left\{ 1 \right\}\)
B.\(D = \mathbb{R}\backslash \left\{ { - 1} \right\}\)
C.\(D = \mathbb{R}\backslash \left\{ { \pm 1} \right\}\)
D.\(D = \mathbb{R}.\)

\(y = \frac{{\sqrt[3]{{x - 1}}}}{{\left| {x - 2} \right| - \left| {x + 2} \right|}}\)
A.\(D = \mathbb{R}.\)
B.\(D = \mathbb{R}\backslash \left\{ 2 \right\}\)
C.\(D = \mathbb{R}\backslash \left\{ { \pm 2} \right\}\)
D.\(D = \mathbb{R}\backslash \left\{ 0 \right\}\)

\(y = \sqrt {x - 1} + \sqrt {9 - {x^2}} \)
A.\(D = \left( {1;3} \right).\)
B.\(D = \left[ { - 3;1} \right]\)
C.\(D = \left[ {1;\,\,3} \right].\)
D.\(D = \left( { - \infty ;3} \right] \cup \left[ {1;3} \right]\)

\(y = \sqrt {x - m} \) xác định trên \(\left( {1; + \infty } \right).\)
A.\(m \ge 1\)
B.\(m \le 1\)
C.\(m \ge 0\)
D.\(m \le 0\)

\(y = \sqrt {x - m} + \sqrt {x - 2m - 2} \) có tập xác định là \(\left[ {0;\,\, + \infty } \right).\)
A.\(m = 1\)
B.\(m \ge 1\)
C.\(m = - 1\)
D.\(m \ge - 1\)