3+3+54+34+32+34+567=?
giai duoc mik cho 3 tik:))
67+39+45+97+28+34+63-43+34+23+98-23-76=??????
Ai đúng mik tik cho ai kbn mik tik cho ai tik cho mik 3 tik mik tik lại cho
Số cần tìm là :
67+39+45+97+28+34+63-43+34+23+98-23-76=386
Đ/s : 386
a)x-(32-27)=3x-(32+x)
b)45-(2x-37)=+x-54
c)42-(3-x)=25-34
d)(25-2x)-34=x-54
36+34+45+46=??
3 người trả lời nhanh nhất mik tik cho nha !!!
TL:
36 + 34 + 45 + 46 = 161
HT~
bằng 161 bạn ơi
34+3 =????
Ai nhanh mik tik nhớ kb nha
2/3+14/15+34/35+....+9899/9900
giúp mik nha
giải ra bài giải
mik sẽ tik cho
=2/1x3+14/3x5+...............................................+9899/
M= 3^2.10^3-[13^2-(5^2+4+2^2+15)]10^3
giup mih vio nha!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
truoc 8 gio ai giai duoc mik se tik cho nguoi do
M=9.1000-[169-(25+4+4+15)]1000
M=9000-[169-48]1000
M=9000-121.1000
M=9000-121000
M=-112000
Ta có \(3^2.10^3-\left[13^2-\left(5^2+4+2^2+15\right)\right]10^3\)
\(=9.1000-\left[169-\left(25+4+4+15\right)\right]1000\)
\(=9.1000-\left(169-48\right)1000\)
\(=1000\left[9-\left(169-48\right)\right]\)
\(=1000.\left(-112\right)\)
\(=-112000\)
A=1+3+32+33+34+..........+32022.Tìm số dư khi chia A cho 13
cứu mik =__)
\(A=1+3+3^2+3^3+...+3^{2022}\)
\(=1+\left(3+3^2+3^3\right)+...+\left(3^{2020}+3^{2021}+3^{2022}\right)\)
\(=1+3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{2020}\left(1+3+3^2\right)\)
\(=1+13\left(3+3^4+...+3^{2020}\right)\)
=>A chia 13 dư 1
chứng minh 1+3+32+33+34+...+32023+32024 chia hết cho 13
giúp mik với !!😥😥😥
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
Cho S = 1+3+32+33+34+35+36+37+38+39.Chứng tỏ rằng S chia hết cho 4
Giup mik vs
\(S=1.\left(1+3\right)+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^8\left(1+3\right)\)
\(S=4x\left(1+3^2+...+3^8\right)\)
Vì 4 chia hết cho 4 nên S chia hết cho 4