Đề kiểm tra 45 phút chương 1: Số hữu tỉ, số thực - Đề số 1

${x^{18}}:{x^6}(x\; \ne 0)$

${x^4}.{\rm{ }}{x^8}$

${\left( {{x^3}} \right)^4}$

$\left| {-{\rm{ }}0,4} \right|{\rm{ }} = -{\rm{ }}0,4$

$\left| {-\,0,4} \right| =  \pm {\rm{ }}0,4$

$\left| {-{\rm{ }}0,4} \right| = 0$

$\dfrac{6}{8}$

$\dfrac{6}{{15}}$

$\dfrac{8}{{15}}$

\(17\)

\(135\)

\(35\)

\(x = 0\)

$x = \;\dfrac{1}{2}$

\(x =  - \dfrac{1}{2}\)

$\dfrac{5}{{15}} = \dfrac{{ - 9}}{{ - 27}};\,\dfrac{{15}}{5} = \dfrac{{ - 27}}{{ - 9}};\,\dfrac{5}{{ - 9}} = \dfrac{{15}}{{ - 27}};\,\dfrac{{ - 9}}{5} = \dfrac{{ - 15}}{{27}}$

$\dfrac{5}{{15}} = \dfrac{{ - 9}}{{ - 27}};\,\dfrac{{15}}{5} = \dfrac{{ - 27}}{9}$

$\dfrac{{15}}{5} = \dfrac{9}{{27}};\,\dfrac{{15}}{5} = \dfrac{{ - 27}}{{ - 9}};\,\dfrac{5}{{ - 9}} = \dfrac{{15}}{{ - 27}};\,\dfrac{{ - 9}}{5} = \dfrac{{ - 15}}{{27}}$

\(33;27\)

\(27;44\)

\(27;34\)

Là số nguyên âm        

Là số nguyên dương

Là số hữu tỉ dương

\(\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} = \dfrac{{x + y + z}}{{a + b + c}}\)

\(\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} = \dfrac{{x - y - z}}{{a - b - c}}\)

\(\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{z}{c} = \dfrac{{x - y + z}}{{a - b + c}}\)

$\dfrac{2}{3} \in Z$

$ - \dfrac{9}{2} \notin Q$

$ - \;6 \in N$

\(x = 5\)

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

\(x = 3\)

$a = 3 - 2k\,\left( {k \in {\mathbb{N}^*}} \right)$

$a = 3 + k\,\left( {k \in {\mathbb{N}^*}} \right)$

$a = 2k\,\left( {k \in {\mathbb{N}^*}} \right)$

$x = 1$

$x = \dfrac{5}{2}$

$x =  - \dfrac{5}{2}$

${x_0} = 1$

${x_0} > 1$    

${x_0} =  - 1$

\(\dfrac{{ - 14}}{5}\)

\(\dfrac{4}{5}\)

\(\dfrac{{14}}{5}\)

$A > 1$          

$A > 2$

$A = 1$

$n = 0$

$n = 3$

$n = 2$

\(x = 1\)          

\(x = 2\)

\(x = 3\)

\(\dfrac{{5a + 3b}}{{5a - 3b}} = \dfrac{{5c + 3d}}{{5c + 3d}}\)

\(\dfrac{{5a - 3b}}{{5a - 3b}} = \dfrac{{5c + 3d}}{{5c - 3d}}\)               

\(\dfrac{{5a - 3b}}{{5a - 3b}} = \dfrac{{5c + 3d}}{{5c - 3d}}\)

\(3\)

\(4\)

\(1\)

\(3\)

\(8\)

\( - 8\)

\(\dfrac{{15}}{{59}}\)

\(\dfrac{{59}}{{15}}\)

\(\dfrac{{15}}{{28}}\)

\(\dfrac{{a_1^3 + a_2^3 + a_3^3}}{{a_2^3 + a_3^3 + a_4^3}} = \dfrac{{{a_4}}}{{{a_1}}}\)

\(\dfrac{{a_1^3 + a_2^3 + a_3^3}}{{a_2^3 + a_3^3 + a_4^3}} = \dfrac{{{a_2}}}{{{a_4}}}\)

\(\dfrac{{a_1^3 + a_2^3 + a_3^3}}{{a_2^3 + a_3^3 + a_4^3}} = \dfrac{{{a_3}}}{{{a_4}}}\)

$5511$

$5151$

$1515$