cho m= 5+5^2+5^3+5^4+...+5^2006;n= \(\frac{1}{4}\).(5^2007-129)
chứng tỏ m-n là số nguyên
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cho m=5+5^2+5^3+5^4+5^2006,n=1/4(5^2007-1290)
chứng tỏ m -n là số tự nhiên
cho M= 5+5^2+5^3+5^4+...+5^2006
N=1/4(5^2007-129)
chứng minh rằng M-N là một số nguyên
\(M=5+5^2+5^3+.......+5^{2006}\)
\(\Leftrightarrow5M=5^2+5^3+.......+5^{2006}+5^{2007}\)
\(\Leftrightarrow5M-M=\left(5^2+5^3+.....+5^{2007}\right)-\left(5+5^2+......+5^{2006}\right)\)
\(\Leftrightarrow4M=5^{2007}-5\)
\(\Leftrightarrow M=\dfrac{5^{2007}-5}{4}\)
Mà \(N=\dfrac{5^{2007}-129}{4}\)
\(\Leftrightarrow M-N=\dfrac{5^{2007}-5}{4}-\dfrac{5^{2007}-129}{4}\)
\(\Leftrightarrow M-N=\dfrac{129}{4}\)
Bạn xem lại có sai đề k ?
S = 5^2 + 5^3 + 5^4 + ... + 5^2006 chia het cho 126
S = 5 + 52 + 53 +....+52006
S= (5+52+53+54+55+56) +.....+ ( 22001+52002+52003+52004+52005+52006)
S= 5 x ( 1+5+52+53+5455 ) +......+ 52001x (1+5+5 2+53+54+55)
S= 5 x 3906+.........+ 52001 x 3906
S = 3906x( 5+..+52001)
S = 3906 x ( 5+...+52001)
S = 126 x 3 x ( 5+...+52001)
=> S chia hết 126
a.(3^5.3^7):3^10+5.2^4-7^3:7
b.3^2[(5^2-3):11]-2^4+2.10^3
c.(6^2007-6^2006):6^2006
d.(5^2001-5^2000):5^2000
e.(7^2005+7^2004):7^2004
f.(5^7+7^5).(6^8+8^6).(2^4-4^2)
g.(7^5+7^9).(5^4+5^6).(3^3.3-9^2)
h.[(5^2.2^3-7^2.2):2].6-7.2^5
mng giúp e vs e c.ơn
Tính tổng
S = 5 + 5^2 + 5^3 + 5^4+....+ 5^2006
Tặng 2 tick cho ai nhanh nhất nhé
5S-S=(5^2+5^3+5^4+...+5^2007)-(5+5^2+5^3+...+5^2006)
4S=5^2007-5
S=(5^2007-5):4
\(S=5+5^2+5^3+5^4+...+5^{2006}\)
\(5S=5^2+5^3+5^4+5^5+...+5^{2007}\)
\(5S-S=\left(5^2+5^3+5^4+5^5+...+5^{2007}\right)-\left(5+5^2+5^3+5^4+...5^{2006}\right)\)
\(4S=5^{2007}-5\)
\(\Rightarrow S=\frac{5^{2007}-5}{4}\)
tính bằng cách hợp lý nhất :
a ) 2006 x 2004 - 2006 x 1004
b) 5/2 x 3/5 + 1/5 + 4/5
a)2006x(2004-1004)
=2006x1000
=2006000
b)=15/10+1
=25/10
=5/2
a ) 2006 x 2004 - 2006 x 1004
=2006x(2004-1004)
=2006x1000
=2006000
b) 5/2 x 3/5 + 1/5 + 4/5
= 95/2 x 3/5) + (1/5 + 4/5)
= 15/10+1
=25/10
=5/2
Tính:
a) (2^2007 + 2^2006) : 2^2006 b) (3^2011 + 3^2010) : 3^2010
c) (5^2001 + 5^2000) : 5^2000 d) (4^2001 + 4^2000) : 4^2000
e) (6^2005 + 6^2004) : 6^2004 f) (7^2011 + 7^2010) : 7^2010
\(a,\left(2^{2007}+2^{2006}\right):2^{2006}=2^{2007}:2^{2006}+2^{2006}:2^{2006}=2+1=3\\ b,\left(3^{2011}+3^{2010}\right):3^{2010}=3^{2011}:3^{2010}+3^{2010}:3^{2010}=3+1=4\\ c,\left(5^{2001}+5^{2000}\right):5^{2000}=5^{2001}:5^{2000}+5^{2000}:5^{2000}=5+1=6\)
Tương tự là d,e,f và kết quả đúng lần lượt là 5,7,8 nha
cho M = 1 + 51 + 52 + ... + 52005 và N = 52006
chứng tỏ rằng M < N phần 4
\(M=1+5+5^2+...+5^{2005}\)
\(\Rightarrow5M=5+5^2+5^3+...+5+5^{2006}\)
\(\Rightarrow5M-M=\left(5+5^2+...+5^{2006}\right)-\left(1+5+...+5^{2005}\right)\)
\(\Rightarrow5M-M=4M=5^{2006}-1\Rightarrow M=\frac{5^{2006}-1}{4}\)
\(\frac{N}{4}=\frac{5^{2006}}{4}>\frac{5^{2006}-1}{4}=M\Rightarrow M< \frac{N}{4}\)
1 6 41 .[ 12+ 12 19 - 12 37 - 12 53 3+ 1 3 - 3 37 - 3 53 : 4+ 4 17 + 4 19 + 4 2006 5+ 5 17 + 5 19 + 5 2006 ]. 124242423 237373735