( 1 - 1/2 ) . ( 1 - 1/3) . ( 1 - 1/4) ... ( 1 - 1/2009) . ( 1 - 1/2010)
Những câu hỏi liên quan
ai đó giúp mk với mk xin chân thành cảm ơn! a=(2010+2010/2+2009/3+2008/4+...+1/2011/ 1/2+1/3+...+1/2011) / (1/2+1/3+1/4+1/5+...+1/2009+1/2010+1/2011)
a) 2010/1+2009/2+2008/3+ ... +1/2010+2010 : 1+1/2+1/3+ ... +1/2010=
b) 1/2011+1/2010+1/2009+ ... +1/3+1/2 : 2010/1+2009/2+2008/3+ ... +1/2010=
a = 1/2 nhân 2 + 1/3 nhân 3 + 1/4 nhân 4 + .....+ 1/2009 nhân 2009 + 1/2010 nhân 2010
so sánh a với 1
a=1/2.2+1/3.3+1/4.4+...+1/2009.2009+1/2010.2010(có 2009 số hạng)
a=1+1+1+...+1+1(2009 số 1)
a=1.2009=2009
Vậy a>1
Đúng 0
Bình luận (0)
https://scratch.mit.edu/projects/782275470
Đúng 0
Bình luận (0)
Cho A = 1/2001+2/2009+3/2008+........2009/+ 2010/1, B = 1+1/2+1/3+1/4+1/5+1/6+.......1/2010+1/2011. Tính A/B
Cho P = 1/2 + 1/3 +1/4 +...+1/2011 + 1/2012
Q = 1/2011 + 2/2010 + 3/2009 +...+ 2009/3 + 2010/2 + 2011/1
Cho A=1×2×3×4×.....×2009×2010×(1+1/2+1/3+1/4+...+1/2009+1/2010)
Chứng minh:A chia hết cho 2011
TÍNH TỔNG : 1/1 : 2 + 1/2 : 3 + 1/3 : 4 + .........+ 1/2009:2010 + 1/2010 : 2011
Hình như đề bài phải là : Tính tổng : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}+\frac{1}{2010.2011}\)
Nếu thế giải như sau : \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}+\frac{1}{2010}-\frac{1}{2011}\)
\(=1-\frac{1}{2011}=\frac{2010}{2011}.\)Vậy tổng đó là 2010/2011.
Đúng 0
Bình luận (0)
Ta có :\(\frac{1}{1}:2+\frac{1}{2}:3+...+\frac{1}{2010}:2011\)
= \(\frac{1}{1}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{3}+...+\frac{1}{2010}\times\frac{1}{2011}\)
= \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2010\times2011}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)
= \(1-\frac{1}{2011}\)
= \(\frac{2010}{2011}\)
Đúng 0
Bình luận (0)
(1/2+1/3-1/4+...+1/2009-1/2010):(1/2006+1/2007+...+1/2010)
Đặt \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(\Rightarrow A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2009}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2010}\right)\)
\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2010}\right)\)
\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2005}\right)\)
\(\Rightarrow A=\frac{1}{2006}+\frac{1}{2007}+...+\frac{1}{2010}\)
\(\Rightarrow\frac{A}{\frac{1}{2006}+\frac{1}{2007}+...+\frac{1}{2010}}=1\)
Đúng 0
Bình luận (0)
Bạn Phạm Tuấn Đạt làm đúng rồi
Dấu \(.\)là dấu nhân
Đặt \(A=\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(B=\frac{1}{2006}+\frac{1}{2007}+...+\frac{1}{2010}\)
Ta có :
\(A=\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(\Rightarrow A=\left(1+\frac{1}{3}+...+\frac{1}{2009}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2010}\right)\)
\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}+\frac{1}{2010}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}...+\frac{1}{2010}\right)\)
\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2005}\right)\)
\(\Rightarrow A=\frac{1}{2006}+\frac{1}{2007}+...+\frac{1}{2010}\)
\(\Rightarrow A=B\)
Nên :
\(\frac{A}{B}=\frac{A}{A}=1\)
Vậy giá trị của biểu thức trên là \(1\)
Đúng 0
Bình luận (0)
1 - 1/2 + 1 /3 - 1/4 + 1/5...........+ 1/2009-1/2010 = 1/1006 + 1007 +....... 1/2010
Ta có :
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2009}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2010}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2009}+\frac{1}{2010}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}...+\frac{1}{2010}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1005}\right)\)
\(=\frac{1}{1006}+\frac{1}{1007}+...+\frac{1}{2010}\)
Đúng 1
Bình luận (0)
tinh tong :
1/1:2+1/2:3+1/3:4+...+1/2009:2010+1/2010:2011