1+ (-2) + 3 (-2) + 2003 + (-2004) + 2005
=1+(3+1).(-2)+(2003+2005)+(-2004)
=1+4.(-2)+4008+(-2004)
=(1+4008)+(-8)+(-2004)
=4009+(-2012)
=1997
1+ (-2) + 3 (-2) + 2003 + (-2004) + 2005
=1+(3+1).(-2)+(2003+2005)+(-2004)
=1+4.(-2)+4008+(-2004)
=(1+4008)+(-8)+(-2004)
=4009+(-2012)
=1997
P=1/2003+1/2004-1/2004 - 2/2002+2/2003-2/2004
5/2003+5/2004-5/2005 3/2002+3/2003-3/2004
Tính: A=(1*2004+2*2003+...+2004*1)/(1*2+2*3+...+2004*2005)
So sánh
a) A= 2003/2004+2004/2005+2005/2003 với 3
b) B= 1/22+1/32+1/42+.......+1/20152 với 1
c) C+ 1/22+1/32+......+1/100 với 75/100
Bài 2. Tính:
a) A = 1 – 2 – 3 + 4 + 5 – 6 – 7 + 8 + ... + 2001 – 2002 – 2003 + 2004.
b) B = 1 + 2 – 3 – 4 + 5 + 6 – 7 – 8 + 9 + ... + 2002 – 2003 – 2004 + 2005 + 2006.Mik sẽ tick cho bạn trả lời nha
Tính: 1/2003+1/2004+1/2005
2/2003+2/2004+2/2005
Bài 1:so sánh: 2017/2018+2018/2019 và ( 2017+2018/2018/2019)
Bài 2: (1/2003+1/2004+1/2005)/(2/2003+2/2004+2/2005)
Bài 3: 2013+ (2013/1+2)+(2013/1+2+30+...+(2013/1+2+3+..+2012)
tính : a)1-2-3+4+5-6-7+8+...+2001-2002-2003+2004
b)1+2-3-4+5+6-7-8+9+...+2002-2003-2004+2005+2006
Bài 1 :
1 . Tính :
\(P=\frac{\frac{1}{2003}+\frac{1}{2004}-\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)
2 . Biết : 13 + 23 + ... + 103 = 3025
Tính : S = 23 + 43 + 63 + .... + 203
A=1-3+5-7+...+2001-2003+2005
B=1-2-3+4+5-6-7+8+...+1993-1994
C=1+2-3-4+5+6-7-8+9+...+2002-2003-2004+2005+2006