a)Ta có : 5\(^5\)- 5\(^4\) + 5\(^3\)

= 5³(5² - 5 + 1 )

=5³ . 21 

Vì 21 \(⋮\)7 nên 21 . 5³\(⋮\)7

Vậy 5⁵ -5⁴ + 5³ \(⋮\)7

 Mấy câu kia b giải tương tự nhé

giải giúp mk với mk sắp đi học rồi

https://hoc247.net/hoi-dap/toan-6/chung-minh-s-1-2-2-2-2-3-2-4-2-5-2-6-2-7-chia-het-cho-3-faq250754.html

S= \(1+2+2^2+...+2^7\)

2S= \(2\cdot\left(2+2^2+...+2^7\right)\)

2S= \(2^1+2^2+...2^8\)

1S= 2S - S = \(\left(2^1+2^2+...2^8\right)-\left(1+2+2^2+...+2^7\right)\)

1S= \(2^1+2^2+...+2^8-1-2-2^2-...-2^7\)

1S= \(2^8-1\)

1S= \(256-1\)

1S= 255

=> 1S chia hết cho 3

Mà 1S= S

=> S chia hết cho 3

Vậy S chia hết cho 3

đó giúp mk đi mà

à, mk quên chưa nói là ai giúp mk sẽ được luôn 2SP đó

giúp mk nha

cảm ơn nhiều!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

những thánh giỏi toán ơi giúp mk được ko

mk năn nỉ đó

bài 1 : thực hiện phép tính

a) 3.5²+15.2²-26:2

= 3.25 + 15.4 - 26 : 2

= 75 + 60 - 13

= 135 - 13

= 122

b) 20:2²+5⁹:5⁸

= 20:4 + 5

= 5 + 5

= 10

c) 100:5²+7.3²

= 100:25 + 7.9

= 4 + 63

= 67

d) 295-(31-2².5)²

= 295-(31-4.5)²

= 295 - 11²

= 295 - 121

= 174

e) (-47)-(45.2⁴-5².12):14

= (-47)-(45.16-25.12):14

f) (-2011)+5.300-(17-7)²

(-2011)+5.(300-10²)

= (-2011)+5.(300-100)

= (-2011)+5.200

= (-2011)+1000

= -1011

g) 5.29-(6-1)²-129

= 5.(29-5²)-129

= 5.(29-25)-129

= 5.4-129

= 20-129

= -109

Đúng thì tik cái nha ! Thanks nhiều !

\(\left(2^{10}+2^9\right)+\left(2^8+2^7\right)+....+\left(2^2+2\right)\)

\(=2^9.\left(2+1\right)+2^7.\left(2+1\right)+...+2.\left(2+1\right)\)

\(=2^9.3+2^7.3+...+2.3\)

\(=3.\left(2^9+2^7+...+2\right)⋮3\)

P/S: mấy bài khác tương tự

\(a,2^{10}+2^9+2^8+...+2\)

\(=\left(2^{10}+2^9\right)+\left(2^8+2^7\right)+...+\left(2^2+2\right)\)

\(=2^9\left(2+1\right)+2^7\left(2+1\right)+...+2\left(2+1\right)\)

\(=2^9.3+2^7.3+...+2.3\)

\(=3\left(2^9+2^7+...+2\right)⋮3\left(đpcm\right)\)

\(b,1+3+3^2+3^3+...+3^{99}\)

\(=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{98}+3^{99}\right)\)

\(=4+3^2\left(1+3\right)+...+3^{98}\left(1+3\right)\)

\(=4+3^2.4+...+3^{98}.4\)

\(=4\left(1+3^2+...+3^{98}\right)⋮4\left(đpcm\right)\)

\(c,1+5+5^2+5^3+...+5^{1975}\)

\(=\left(1+5\right)+\left(5^2+5^3\right)+...+\left(5^{1974}+5^{1975}\right)\)

\(=6+5^2\left(1+5\right)+...+5^{1974}\left(1+5\right)\)

\(=6+5^2.6+...+5^{1974}.6\)

\(=6\left(1+5^2+...+5^{1974}\right)⋮6\left(đpcm\right)\)

Câu a:

(x - 5)^2 - 2x - 2^4 = 2x

(x - 5)(x - 5) - 2x - 16 - 2x = 0

x^2 - 5x - 5x + 25 - 2x - 16 - 2x = 0

x^2 - (5x + 5x +2x + 2x) + (25 - 16) = 0

x^2 - 14x + 9 = 0

(x^2 - 7x) - (7x - 49) - 40 = 0

x(x - 7) - 7(x - 7) - 40 = 0

(x - 7)(x - 7) - 40 = 0

(x - 7)^2 = 40

x - 7 = \(\sqrt{40}\) hoặc x - 7 = -\(\sqrt{40}\)

x - 7 = - \(\sqrt{40}\)

x = 7 - \(\sqrt{40}\)

x - 7 = \(\sqrt{40}\)

x = 7+ \(\sqrt{40}\)

Vậy x ∈ {7 - \(\sqrt{40}\) ; 7+ \(\sqrt{40}\))

Câu b:

3^(x -1) - 7^2 = 2^5 + 0^3

3^(x -1) - 49 = 32 + 0

3^(x - 1) - 49 = 32

3^(x -1) = 32 + 49

3^(x -1) = 81

3^(x-1) = 3^4

x - 1 = 4

x = 4 + 1

x = 5

Vậy x = 5

a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)

\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)

=>x=10

b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)

\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)

\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)

hay \(x\in\left\{0;1;2\right\}\)

c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)

\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)

\(\Leftrightarrow6-x=0\)

hay x=6