Lũy thừa của một số hữu tỉ

${\left( {\frac{{ - 3}}{7}} \right)^3} = \left( {\frac{{ - 3}}{7}} \right).\left( {\frac{{ - 3}}{7}} \right).\left( {\frac{{ - 3}}{7}} \right) = \frac{{( - 3).( - 3).( - 3)}}{{7.7.7}} = \frac{{ - 27}}{{343}}$

Ta có ${a^m}.{a^n} = {a^{m + n}}$, ${\left( {a.b} \right)^m} = {a^m}.{b^m}$ và ${\left( {{a^m}} \right)^n} = {a^{m.n}}$ nên  C sai.

Ta có: 9⁴ . 3⁵ = (3²)⁴ . 3⁵ = 3^2.4 . 3⁵ = 3⁸ . 3⁵ = 3⁸⁺⁵ = 3¹³

$8:{\left( {\dfrac{2}{3} - \dfrac{3}{4}} \right)^2} = 8:{\left( {\dfrac{8}{{12}} - \dfrac{9}{{12}}} \right)^2}$$= 8:{\left( {\dfrac{{ - 1}}{{12}}} \right)^2} = 8:\dfrac{1}{{144}} = 8.144$$= 1152$

Ta có ${x^1} = x;$${x^0} = 1$$\left( {x \ne 0} \right)$ nên A, B, C sai

${\left( {\dfrac{x}{y}} \right)^n} = \dfrac{{{x^n}}}{y^n}\left( {y \ne 0;\,n \in \mathbb{N}} \right)$ nên D đúng.

+) (-4)³ . 4⁵ = - 4³ . 4⁵ = - 4³⁺⁵ = - 4⁸

Vậy A sai

+) a^m : aⁿ = a^m-n

Vậy B sai

+) (-6)²⁰²¹ = - 6²⁰²¹ ( vì 2021 là số lẻ)

Vậy C sai

+) [(-3)²]⁵ = (3²)⁵ = 3^2.5 = 3¹⁰

Vậy D đúng

27^x . 3⁴ = 9⁵

$\Rightarrow$ (3³)^x . 3⁴ = (3²)⁵

$\Rightarrow$3^3.x . 3⁴ = 3¹⁰

$\Rightarrow$3^3x =  3¹⁰ : 3⁴

$\Rightarrow$3^3x = 3⁶

$\Rightarrow$3x = 6

$\Rightarrow$x = 2

Vậy x = 2

Ta có ${\left( {-2019} \right)^0} = 1$ nên A đúng.

+)  ${4^6}:{\rm{ }}{4^4} = {4^2} = 16$ nên C đúng

+)  ${\left( {-3} \right)^3}.{\left( {-{\rm{ }}3} \right)^{{\rm{ }}2}} = {\left( { - 3} \right)^{3 + 2}} = {\left( { - 3} \right)^5}$nên D đúng

+) $\left( {0,5} \right).{\left( {0,5} \right)^2} = {\left( {0,5} \right)^3} = {\left( {\dfrac{1}{2}} \right)^3} = \dfrac{1}{8}$ nên B sai.

Ta có: A = 1 + 3 +  3² +…+ 3²⁰²²

$\Rightarrow$3.A = 3. ( 1 + 3 +  3² +…+ 3²⁰²²) = 3 + 3² +  3³ +…+ 3²⁰²³

$\Rightarrow$ 3. A – A = 3 + 3² +  3³ +…+ 3²⁰²³ – (1 + 3 +  3² +…+ 3²⁰²²)

2A = 3 + 3² +  3³ +…+ 3²⁰²³ – 1 - 3 -  3² - …- 3²⁰²² = 3²⁰²³ – 1

$\Rightarrow A = \frac{{{3^{2023}} - 1}}{2}$

Ta có: ${2^{24}} = {2^{3.8}} = {\left( {{2^3}} \right)^8} = {8^8}$

Khối lượng của Mặt Trời gấp khoảng số lần khối lượng Trái Đất là:

$\frac{{1{\rm{ }}988{\rm{ }}550{\rm{ }}.{\rm{ }}{{10}^{21}}}}{{0,{{6.10}^{22}}}} = \frac{{1{\rm{ }}988{\rm{ }}550{\rm{ }}.{\rm{ }}{{10}^{21}}}}{{{{6.10}^{21}}}} = 331425$ ( lần)

(2x+1)³ – 1 = -344

$\Leftrightarrow$(2x+1)³ = -344 + 1

$\Leftrightarrow$(2x+1)³ = -343

$\Leftrightarrow$(2x+1)³  = (-7)³

$\Leftrightarrow$2x + 1 = -7

$\Leftrightarrow$2x = -7 – 1

$\Leftrightarrow$2x = -8

$\Leftrightarrow$x = -4

Vậy x = -4

$a{\rm{ }}:{\left( {\dfrac{1}{3}} \right)^4} = {\left( {\dfrac{1}{3}} \right)^3}$

$a = {\left( {\dfrac{1}{3}} \right)^3}.{\left( {\dfrac{1}{3}} \right)^4}$

$a = {\left( {\dfrac{1}{3}} \right)^{3 + 4}}$

$a = {\left( {\dfrac{1}{3}} \right)^7}$

Thay x = 3 vào M ta được:

$\begin{array}{l}M = \frac{{ - {x^2} + 2x - 1}}{{{{(2x)}^3}}}\\ = \frac{{ - {3^2} + 2.3 - 1}}{{{{(2.3)}^3}}}\\ = \frac{{ - 9 + 6 - 1}}{{{6^3}}}\\ = \frac{{ - 4}}{{216}}\\ = \frac{{ - 1}}{{54}}\end{array}$

${20^n}\;:\;{5^n} = 4$  

${(20:5)^n} = 4$

${4^n} = 4$

$n = 1$

$A = \dfrac{{{2^7}{{.9}^3}}}{{{6^5}{{.8}^2}}} = \dfrac{{{2^7}.{{\left( {{3^2}} \right)}^3}}}{{{2^5}{{.3}^5}.{{\left( {{2^3}} \right)}^2}}} = \dfrac{{{2^7}{{.3}^6}}}{{{2^5}{{.2}^6}{{.3}^5}}}$$= \dfrac{{{2^7}{{.3}^6}}}{{{2^{11}}{{.3}^5}}} = \dfrac{{1.3}}{{{2^4}.1}} = \dfrac{3}{{16}}$

Ta có $\dfrac{{{4^6}{{.9}^5} + {6^9}.120}}{{{8^4}{{.3}^{12}} - {6^{11}}}} = \dfrac{{{{\left( {{2^2}} \right)}^6}.{{\left( {{3^2}} \right)}^5} + {6^9}.120}}{{{{\left( {{2^3}} \right)}^4}{{.3}^{12}} - {6^{11}}}}$$= \dfrac{{{2^{12}}{{.3}^{10}} + {6^9}.6.20}}{{{2^{12}}{{.3}^{12}} - {6^{11}}}} = \dfrac{{{2^2}{{.2}^{10}}{{.3}^{10}} + {6^{10}}.20}}{{{{\left( {2.3} \right)}^{12}} - {6^{11}}}}$$= \dfrac{{{2^2}{{.6}^{10}} + {6^{10}}.20}}{{{6^{12}} - {6^{11}}}}$$= \dfrac{{{6^{10}}\left( {{2^2} + 20} \right)}}{{{6^{10}}\left( {{6^2} - 6} \right)}} = \dfrac{{24}}{{30}} = \dfrac{4}{5}$

${\left( {5x - 1} \right)^6} = 729$

${\left( {5x - 1} \right)^6} = {(3)^6}$

Trường hợp 1:

$\begin{array}{l}5x-1 = 3\\5x = 4\\x = \dfrac{4}{5}\end{array}$

Trường hợp 2:

$\begin{array}{l}5x-1 =  - 3\\5x =  - 2\\x =  - \dfrac{2}{5}\end{array}$

Vậy $x = \dfrac{4}{5}$ hoặc $x =  - \dfrac{2}{5}$

${\left( {2x + 1} \right)^3} =  - {0,1^3} = {\left( { - 0,1} \right)^3}$

$2x + 1 =  - 0,1$

$2x =  - 0,1 - 1$

$2x =  - 1,1$

$x =  - 1,1:2$

$x =  - 0,55$  

Vậy $x =  - 0,55$.

${5^n} + {5^{n + 2}} = 650$

${5^n} + {5^n}{.5^2} = 650$

${5^n}\left( {1 + {5^2}} \right) = 650$

${5^n}\left( {1 + 25} \right) = 650$

${5^n}.26 = 650$

${5^n} = 650:26$

${5^n} = 25$

${5^n} = {5^2}$

$n = 2$

Vậy $n = 2$