chung minh : 1/5^3+1/6^3+1/7^3+........+1/2004^3 <1/40
Những câu hỏi liên quan
chung minh rang 1/5^3+1/6^3+1/7^3+..........+1/2004^3<1/40
chung minh : 1/5^3+1/6^3+1/7^3+.........+1/2004^3 < 1/40
Chứng minh rằng : 1/65 < 1/5^3 + 1/6^3 + 1/7^3 + ... + 1/2004^3 <1/40
a)chung minh A= 2^1+2^2+2^3+2^4+...2^2010chia het cho 3
b)chung minh B= 3^1+3^2+3^3+3^4+...3^2010chia het cho 4
c)chung minh C= 5^1+5^2+5^3+5^4+...5^2010chia het cho 6
d)chung minh D= 7^1+7^2+7^3+7^4+...7^2010chia het cho 8
a) A=21+22+23+...+22010
A=(21+22)+(23+24)+.....+(22009+22010)
A=(21x3)+(23x3)+.....+(22009x3)
A=3x(21+23+.......+22009)
Vậy A chia hết cho 3.
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A=1+3+3^2+3^3+3^4+3^5+3^6+3^7 chung minh A=(3^8-1):2
\(A=1+3+...+3^7\\ \Rightarrow3A=3+3^2+...+3^8\\ \Rightarrow3A-A=\left(3+3^2+...+3^8\right)-\left(1+3+...+3^7\right)\\ \Rightarrow2A=3^8-1\\ \Rightarrow A=\dfrac{3^8-1}{2}\)
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\(A=1+3+3^2+...+3^7\)
\(3A=3+3^2+3^3+...+3^8\)
\(3A-A=\left(3+3^2+3^3+...+3^8\right)-\left(1+3+3^2+...+3^7\right)\)
\(2A=3^8-1\)
\(\Rightarrow A=\dfrac{3^8-1}{2}\) (đpcm)
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a) 1 - 2 - 3 + 4 +5 - 6 - 7 + ..... + 2001 - 2002 -2003 + 2004
b) 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ..... + 2001 + 2002 - 2003 - 2004
a) \(1-2-3+4+5-6-7+...+2001-2002-2003+2004\)
\(=\left(1-2-3+4\right)+\left(5-6-7+8\right)+...+\left(2001-2002-2003+2004\right)\)
\(=0+0+...+0=0\)
b) \(1+2-3-4+5+6-7-8+...+2001+2002-2003-2004\)
\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(2001+2002-2003-2004\right)\)
\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)
\(=\left(-4\right)\cdot501=\left(-2004\right)\)
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Tính giá trị biểu thức sau :
A = ( 6 : 3/5 - 7/6 * 6/7 ) : ( 21/5 * 10/11 + 57/11 )
B = 59/10 : 3/2 - ( 7/3 * 9/2 - 2 * 7/3 ) : 7/4
C = ( 1 - 1/2 ) * ( 1- 1/3 ) * ( 1 - 1/4 ) * ( 1- 1/5 ) .............. ( 1 - 1/2003 ) * ( 1 - 1/2004 )
A = ( 6 : 3/5 - 7/6 * 6/7 ) : ( 21/5 * 10/11 + 57/11 )
A = ( 10 - 1 ) : ( 42/11 + 57/11)
A = 9 : 9
A = 1
B = 59 /10 : 3/2 - ( 7/3 * 9/2 - 2 * 7/3 ) : 7/4
B = 59/15 - ( 21/2 - 14/3 ) : 7/4
B = 59/15 - 35/6 : 7/4
B = 59/15 - 10/3
B = 3/5
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Xem thêm câu trả lời
số 2 : tính nhanhA 99 - 97 + 95 -93 + 91 - 89 + .......... + 7 - 5 + 3 - 1B 50 - 49 + 48 - 47 + 46 - 45 + ........... + 4 - 3 + 2 - 1C 100 + 98 + 96 + ......... + 2 - 97 - 95 - ......... - 1D 1 + 3 + 5 + 7 + ......... + 999E 1 + 11 + 21 + 31 + ......... + 991F 3 + 7 + 11 + 15 + ......... + 99H 1 + 2 + 3 - 4 - 5 - 6 + 7 + 8 + 9 - 10 - 11 - 12 + .......... + 97 + 98 + 99 - 100 -101 - 102 I 1 - 3 + 5 5 - 7 + 9 - 11 + ... + 2004 - 2007K -1 + 2 - 3 + 4 - 5 + 6 - 7 +.... + 2004 - 2005G 1 - 2...
Đọc tiếp
số 2 : tính nhanh
A = 99 - 97 + 95 -93 + 91 - 89 + .......... + 7 - 5 + 3 - 1
B = 50 - 49 + 48 - 47 + 46 - 45 + ........... + 4 - 3 + 2 - 1
C = 100 + 98 + 96 + ......... + 2 - 97 - 95 - ......... - 1
D = 1 + 3 + 5 + 7 + ......... + 999
E = 1 + 11 + 21 + 31 + ......... + 991
F = 3 + 7 + 11 + 15 + ......... + 99
H = 1 + 2 + 3 - 4 - 5 - 6 + 7 + 8 + 9 - 10 - 11 - 12 + .......... + 97 + 98 + 99 - 100 -101 - 102
I = 1 - 3 + 5 5 - 7 + 9 - 11 + ... + 2004 - 2007
K = -1 + 2 - 3 + 4 - 5 + 6 - 7 +.... + 2004 - 2005
G =1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 +..... + 2004 + 2005
N = 1 - 4 + 7 - 10 + ..... + 2995 - 2998
chứng minh rằng
\(\frac{1}{5^3}+\frac{1}{6^3}+\frac{1}{7^3}+...+\frac{1}{2004^3}\)<\(\frac{1}{40}\)
