Bài 7: Lũy thừa với số mũ tự nhiên. Nhân hai lũy thừa cùng cơ số. Luyện tập

a: \(A=1+2+2^2+...+2^{2023}\)

=>\(2A=2+2^2+2^3+...+2^{2024}\)

=>\(2A-A=2^{2024}+2^{2023}+...+2^2+2-2^{2023}-2^{2022}-...-2^2-2-1\)

=>\(A=2^{2024}-1\)

b: \(A=\left(1+2\right)+2^2+2^3+...+2^{2023}\)

\(=3+2^2\left(1+2\right)+...+2^{2022}\left(1+2\right)\)

\(=3\left(1+2^2+...+2^{2022}\right)⋮3\)

a: \(20-\left[30-\left(5-1\right)^2\right]\)

\(=20-\left[30-4^2\right]\)

\(=20-14=6\)

b: \(71+\dfrac{50}{5+3\left(57-6\cdot7\right)}\)

\(=71+\dfrac{50}{5+3\cdot\left(57-42\right)}\)

\(=71+\dfrac{50}{5+3\cdot15}=71+\dfrac{50}{50}=72\)

c: \(4\cdot\left\{270:\left[50-\left(2^5+45:5\right)\right]\right\}\)

\(=4\cdot\left\{270:\left[50-32-9\right]\right\}\)

\(=4\cdot\left\{\dfrac{270}{50-41}\right\}=4\cdot\dfrac{270}{9}=4\cdot30=120\)

d: \(411-\left[\dfrac{\left(107+3\right)}{5}-2^2\right]\)

\(=411-\left[\dfrac{110}{5}-4\right]\)

=410-22+4

=410-18

=392

e: \(450-5\left[3^2\left(7^5:7^3-41\right)-12\right]+18\)

\(=450-5\left[9\cdot\left(7^2-41\right)-12\right]+18\)

\(=450-5\cdot\left[9\cdot8-12\right]+18\)

=468-5*60

=468-300

=168

f:

\(102-150:\left[18-2\cdot\left(10-8\right)^2\right]+1018^0\)

\(=102-150:\left[18-2\cdot4\right]+1\)

\(=103-\dfrac{150}{18-8}=103-15=88\)

Đặt \(A=1+3+3^2+3^3+3^4+\cdot\cdot\cdot+3^{2023}+3^{2024}\)

\(=(1+3+3^2)+(3^3+3^4+3^5)+(3^6+3^7+3^8)+\dots+(3^{2022}+3^{2023}+3^{2024})\\=13+3^3\cdot(1+3+3^2)+3^6\cdot(1+3+3^2)+\dots+3^{2022}\cdot(1+3+3^2)\\=13+3^3\cdot13+3^6\cdot13+\dots+3^{2022}\cdot13\\=13\cdot(1+3^3+3^6+\dots+3^{2022})\)

Vì \(13\cdot(1+3^3+3^6+\dots+3^{2022})\vdots13\)

nên \(A\vdots13\)

\(\Rightarrowđpcm\)

Đặt S=1+3+3²+3³+3⁴+⋅⋅⋅+3²⁰²³+3²⁰²⁴

S=(1+3+3²)+(3³+3⁴+3⁵)+⋯+(3²⁰²²+3²⁰²³+3²⁰²⁴)

S=13+3³(1+3+3²)+...+3²⁰²²(1+3+3²)

S=13+3³.13+...+3²⁰²².13

S=13(3³+...+3²⁰²²) ⋮ 13

Vậy S⋮13

$(2x+3).{}^{}=351$

$\Rightarrow (2x+3).27=351$

$\Rightarrow 2x+3=351:27$

$\Rightarrow 2x+3=13$

$\Rightarrow 2x=10$

$\Rightarrow x=10:2$

$\Rightarrow x=5$

Vậy, $x=5$

Bạn muốn làm gì với biểu thức này nhỉ?

\(3+3^3+3^5+3^7+...+3^{31}\)

\(=\left(3+3^3\right)+\left(3^5+3^7\right)+...+\left(3^{29}+3^{31}\right)\)

\(=\left(3+3^3\right)+3^4\left(3+3^3\right)+...+3^{28}\left(3+3^3\right)\)

\(=30\cdot\left(1+3^4+...+3^{28}\right)⋮30\)

\(\left(125^3.7^5-175^5:5\right):2016^{2017}\)

= \(\left[\left(5^3\right)^3.7^5-\left(5^2.7\right)^5:5\right]:2016^{2017}\)

= \(\left[5^{3.3}.7^5-5^{2.5}.7^5:5\right]:2016^{2017}\)

= \(\left[5^9.7^5-5^{10}.7^5:5\right]:2016^{2017}\)

= \(\left[5^9.7^5-5^9.7^5\right]:2016^{2017}\)

= \(0:2016^{2017}\)

= \(0\)

\(\dfrac{125^3\cdot7^5-175^5:5}{2016^{2017}}\)

\(=\dfrac{5^9\cdot7^5-5^{10}\cdot7^5:5}{2016^{2017}}\)

\(=\dfrac{5^9\cdot7^5-5^9\cdot7^5}{2016^{2017}}\)

=0

Gọi số tự nhiên cần tìm là x

Theo đề, ta có: \(x^2\in\left\{50;51;...;70\right\}\)

mà x nguyên

nên \(x^2=64\)

=>x=8

lập phương của 8 là \(8^3=512\)

yêu cầu là j vậy bạn

A = 2 + 22 + 23 + … + 22004 . Chứng minh rằng A chia hết cho 3 , cho 7.