Tính 4/2*4+4/4*6+...+4/2008*2010
Tính
A=1-2+3+4-5-6+7+8-9-.....+2007+2008-2009-2010
B=2^2005-2^2003-2^2001-......-2-1
C=2+4+4+8+10-12-14+16+......-2006+2008+2010-2012
D=1-2-3-4+5-6-7-8+9-10-11-......-2010
4/2*4+4/4*6+4/6*8+...+4/2008*2010
4/2.4 + 4/4.6 + 4/6.8 + ... + 4/2008.2010
= 2.(1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ... + 1/2008 - 1/2010)
= 2.(1/2 - 1/2010)
= 2.502/1005
= 1004/1005
\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(=2.\frac{502}{1005}\)
\(=\frac{1004}{1005}\)
\(\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(=2\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{2010}\right)=2\left(\frac{1005}{2010}-\frac{1}{2010}\right)=2.\frac{1004}{2010}\)
\(=\frac{1004}{1005}\)
Tính tổng:
a,S1=1+(-2)+3+(-4)+..........+2009+(-2010)
b,S2=1+(-2)+(-3)+4+5+(-6)+(-7)+............+2008+2009+(-2010)
a,S1=1+(-2)+3+(-4)+..........+2009+(-2010)
S1=-1.(2010:2)
S1=-1005
b,S2=1+(-2)+(-3)+4+5+(-6)+(-7)+............+2008+2009+(-2010)
S2=-1.(2010:2)
S2=-1.1005
S2=-1005
C = 4 / 2 nhan 4 + 4 / 4 nhan 6 + 4 / 6 nhan 8 + ... + 4 / 2008 nhan 2010
\(C=\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...+\frac{4}{2008\cdot2010}\)
\(C=\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+...+\frac{2}{1004\cdot1005}\)
\(C=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1004}-\frac{1}{1005}\)
\(C=1-\frac{1}{1005}\)
\(C=\frac{1004}{1005}\)
Tính tổng sau
A=1-2+3-4+5-....-2008+2009
B=1+2-3-4+5+6-7-...-2007-2008+2009+2010
Bài làm:
\(A=1-2+3-4+5-...-2008+2009\)
\(A=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...+\left(2007-2008\right)+2009\)
\(A=-1-1-1-...-1+2009\)(1004 số -1)
\(A=-1004+2009=1005\)
\(B=1+2-3-4+5+6-7-...-2007-2008+2009+2010\)
\(B=1+\left(2-3-4+5\right)+\left(6-7-8+9\right)+...+\left(2006-2007-2008+2009\right)+2010\)
\(B=1+0+0+...+0+2010\)
\(B=2011\)
Học tốt!!!!
Tính nhanh
C= 4 phần 2 nhân 4 + 4 phan 4 nhân 6+ 4 phần 6 nhân 8+........+4 phần 2008 nhân 2010
\(C=\frac{4}{2\times4}+\frac{4}{4\times6}+\frac{4}{6\times8}+...+\frac{4}{2008\times2010}\)
\(C=2\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(C=2\times\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(C=2\times\frac{502}{1005}\)
\(C=\frac{1004}{1005}\)
\(C=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(C=2\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{2010-2008}{2008.2010}\right)\)
\(C=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(C=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(C=\frac{1004}{1005}\)
\(C=\frac{4}{2\cdot4}+\frac{4}{4\cdot6}+\frac{4}{6\cdot8}+...+\frac{4}{2008\cdot2010}\)
\(C=2\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+...+\frac{2}{2008\cdot2010}\right)\)
\(C=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(C=2\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(C=2\cdot\frac{1004}{2010}\)
\(C=\frac{1004}{1005}\)
tính tổng : 0-2+4-6+8-10+12-14+...+2008-2010
nhóm 2 số đầu rối lại nhóm 2 số tiếp theo cứ thế thôi
ko rảnh mà viết hẳn bài ra
sorry nhóe
k nha
Gọi A=0-6-10-14-....-2010
A=0-(6+10+14+....+2010)
số số hạng (2010-6):4+1=502
tổng (2010+6).502:2=506016
A=0-506016
A=-506016
Gọi B=4+8+12+....+2008
số số hạng (2008-4):4+1=502
tổng (2008+4).502:2=505012
-506016+505012=-1004
Vậy -1004
Tính toorng :1+2-3-4+5+6-7-8+9+...+2006-2007-2008+2009+2010
=1+(2-3-4+5)+(6-7-8+9)+...+(2006-2007-2008+2009)+2010
=1+0+0+...+0+2010
=2011
Ta thấy tổng của 4 số bắt đầu từ 2 thì đều =0 (2-3-4+5=0,6-7-8+9=0)Ta đặt A=\(1+2-3-4+5+6-7-8+9+...+2006-2007-2008+2009+2010\)
= \(1+\left(2-3-4+5\right)+\left(6-7-8+9\right)+...+\left(2006-2007-2008+2009\right)+2010=1+0+0+...+0+2010=2011\)
Tính A=1+2-3-4+5+6-7-8+....-2007-2008+2009+2010
A=2011 nha