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bài 1) a) \(1+2+3+4+........+2005+2006\)
\(\Leftrightarrow\) \(\left(1+2006\right)+\left(2+2005\right)+........+\left(1003+1004\right)\)
\(\Leftrightarrow\) \(2007.\dfrac{2006}{2}=2007.1003=2013021\)
b) \(5+10+15+.......+2000+2005\)
\(\Leftrightarrow\) \(\left(2005+5\right)\left(2000+10\right)+.......+\left(1000+1010\right)\)
\(\Leftrightarrow\) \(2010.\dfrac{2005}{5}=2010.401=405010\)
c) \(140+136+132+.......+64+60\)
\(\Leftrightarrow\) \(\left(140+60\right)+\left(136+64\right)+.......+\left(100+100\right)\)
\(\Leftrightarrow\) \(200.10\) = \(2000\)
1)
a) \(1+2+3+4+.....+2005+2006\)
Số các số hạng của dãy trên là:
\((2006-1):1+1=2006\)
Tổng dãy là:
\(\dfrac{2006\left(2006+1\right)}{2}=2013021\)
b) \(5+10+15+.....+2000+2005\)
Số các số hạng của dãy là:
\((2005-5):5+1=401\)
Tổng dãy là:
\(\dfrac{401\left(2005+5\right)}{2}=403005\)
c)\(140+136+132+.....+64+60\)
\(=60+64+.....+132+136+140\)
Số số hạng của dãy là:
\((140-60):4+1=11\)
Tổng dãy là:
\(\dfrac{11\left(60+140\right)}{2}=1100\)
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\(a,4^{21}:16^5.\)
\(=4^{21}:\left(4^2\right)^5.\)
\(=4^{21}:4^{10}.\)
\(=4^{21-10}=4^{11}.\)
Vậy.....
\(b,32^8:4^{19}.\)
\(=\left(2^5\right)^8:\left(2^2\right)^{19}.\)
\(=2^{40}:2^{38}.\)
\(=2^{40-38}.\)
\(=2^2=4.\)
Vậy.....
\(c,27^{15}:9^{22}.\)
\(=\left(3^3\right)^{15}:\left(3^2\right)^{22}.\)
\(=3^{45}:3^{44}.\)
\(=3^{45-44}.\)
\(=3^1=3.\)
Vậy.....
\(d,25^{10}:125^6.\)
\(=\left(5^2\right)^{10}:\left(5^3\right)^6.\)
\(=5^{20}:5^{18}.\)
\(=5^{20-18}.\)
\(=5^2=25.\)
Vậy.....
~ Hok tốt!!! ~ :))
a, 421 : 165
= 421 : (42 )5
= 421 : 410
= 411
b, 328 : 419
= (25)8 : (22 )19
= 240 : 238
= 22
c, 2715 : 922
= (33 ) 15 : (32 )22
= 345 : 344
= 3
d, 2510 : 1256
= (52)10 : (53)6
= 520 : 518
= 52
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a)n+3\(⋮\)n
n\(⋮\)n
n+3-n\(⋮\)n
3\(⋮\)n
\(\Rightarrow\)n={1,3}
b)7n+8\(⋮\)n
7n\(⋮\)n
7n+8-7n\(⋮\)n
8\(⋮\)n
\(\Rightarrow\)n={1,2,4,8}
c)35-12n\(⋮\)n
12n\(⋮\)n
35-12n-12n\(⋮\)n
35\(⋮\)n
\(\Rightarrow\)n={1,5,7,35}
d)n+8\(⋮\)n+3
n+3\(⋮\)n+3
n+8-(n+3)\(⋮\)n+3
n+8-n-3\(⋮\)n+3
5\(⋮\)n+3
\(\Rightarrow\)n+3={1,5}
\(\Rightarrow\)n={-1,2}
vi x\(\in\)N nen x =2
d)16-3n\(⋮\)n+4
3(n+4)\(⋮\)n+4
16-3n-3(n+4)\(⋮\)n+4
16-3n-3n-12\(⋮\)n+4
4\(⋮\)n+4
\(\Rightarrow\)n+4={1,4}
voi n+4=1\(\Rightarrow\)n=khong tim duoc
voi n+4=4\(\Rightarrow\)n=0
vay n=0
a) n + 3 chia hết cho n
(n chia hết cho n + 3 ) chia hết cho n
=> 3 chia hết cho n
=> n E Ư(3)={ 1;3}
Các câu còn lại bạn tự giải nhé
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1. A = (-2)(-3) - 5.|-5| + 125.\(\left(-\dfrac{1}{5}\right)^2\)
= 6 - 25 + 125.\(\dfrac{1}{25}\)
= -19 + 5
= -14
@Shine Anna
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A = \(\dfrac{1}{4}.\dfrac{7}{3}.12\)
= \(\dfrac{1.7.12}{4.3}\)
= \(7\)
@Nguyễn Thành Đăng
B = \(\dfrac{3}{8}.56.\dfrac{25}{7}.\left(-4\right)\)
= \(-\dfrac{3.56.25.4}{8.7}\)
= -3.100
= -300
@Nguyễn Thành Đăng
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\(ab-ac+bc-c^2+1=0\\ \Rightarrow a\left(b-c\right)+c\left(b-c\right)=-1\\ \Rightarrow\left(a+c\right)\left(b-c\right)=-1\\ \Rightarrow a+c+b-c=0\\ \Rightarrow a+b=0\)
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Cho e hỏi cái này. Ở câu 1 ý, cuối đề là \(-\frac{1}{7}\) sao xuống dưới phải đổi thành -1 thế ạ ? E chưa hiểu lắm :<
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a) C = 1 + 52 + 54 + ... + 52018
\(\Rightarrow\) 25C = 5 + 54 + 56 + ... + 52020
\(\Rightarrow\) 25C - C = (5 + 54 + 56 + ... + 52020) - (1 + 52 + 54 + ... + 52018)
\(\Rightarrow\) 24C = 52020 - 1
\(\Rightarrow\) C = \(\dfrac{5^{2020}-1}{24}\)
b) D = 2 . 4 + 4 . 6 + 6 . 8 + ... + 2016 . 2018
\(\Rightarrow\) 6D = 2 . 4 . 6 + 4 . 6 . (8 - 2) + 6 . 8 . (10 - 4) + ... + 2016 . 2018 . (2020 - 2014)
\(\Rightarrow\) 6D = 2 . 4 . 6 + 4 . 6 . 8 - 4 . 6 . 2 + 6 . 8 . 10 - 6 . 8 . 4 + ... + 2016 . 2018 . 2020 - 2016 . 2018 . 2014
\(\Rightarrow\) 6D = 2016 . 2018 . 2020
\(\Rightarrow\) D = 336 . 2018 . 2020
\(\Rightarrow\) D = 1 369 656 960
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2. \(\left(2,7x-1\dfrac{1}{2}x\right):\dfrac{2}{7}=\dfrac{-21}{4}\)
\(\Leftrightarrow x.\left(\dfrac{27}{10}+\dfrac{-3}{2}\right)=\dfrac{-21}{4}.\dfrac{2}{7}\)
\(\Leftrightarrow x.\left(\dfrac{27}{10}+\dfrac{-15}{10}\right)=\dfrac{-3}{2}\)
\(\Leftrightarrow x.\dfrac{6}{5}=\dfrac{-3}{2}\)
\(\Leftrightarrow x=\dfrac{-3}{2}:\dfrac{6}{5}\)
\(\Leftrightarrow x=\dfrac{-3}{2}.\dfrac{5}{6}\)
\(\Leftrightarrow x=\dfrac{-5}{4}\)
3.\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=1\\2x-\dfrac{3}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1+\dfrac{3}{4}\\2x=\left(-1\right)+\dfrac{3}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{7}{3}\\2x=\dfrac{-7}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}.\dfrac{1}{2}\\x=\dfrac{-7}{3}.\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=\dfrac{-7}{6}\end{matrix}\right.\)
vậy \(x\in\left\{\dfrac{7}{6};\dfrac{-7}{6}\right\}\)
kết quả = -14