tinh nhanh
A=2^2009-2^2008-2^2007-.....-2^2-2^1-1
Tinh nhanh A=2^2009-2^2008-2^2007-.....-2^2-2^1-1
A = 22009 - 22008 - 22007 - ... - 22 - 2 - 1
A = 22009 - (22008 + 22007 + ... + 22 + 2 + 1)
Đặt B = 22008 + 22007 + ... + 22 + 2 + 1
2B = 22009 + 22008 + ... + 23 + 22 + 2
2B - B = (22009 + 22008 + ... + 23 + 22 + 2) - (22008 + 22007 + ... + 22 + 2 + 1)
B = 22009 - 1
=> A = 22009 - (22009 - 1) = 22009 - 22009 + 1 = 0 + 1 = 1
Tính A= 2009/2+2008/(2^2)+2007/(2^3)+...+3/(2^2007)+2/(2^2008)+1/(2^2009)
A=(2008/1+2007/2+...........+2/2007+2/2008)/(1/2+1/3+........+1/2008+1/2009)
2008+2007/2+2006/3+...+2/2007+1/2008
1/2+1/3+1/4+...1/2008+1/2009
so sánh 2008 với tổng 2009 số hạng sau\(s=\frac{2008+2007}{2009+2008}+\frac{^{2008^2+2007^2}}{2009^2+2008^2}+.....+\frac{2008^{2009}+2007^{2009}}{2009^{2009}+2008^{2009}}\)
2008/1+2007/2+2006/3+....+2/2007+1/2008
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1/2+1/3+1/4+.....+1/2008+1/2009
P/s : Lớp 6 nhé bạn
Dấu \(.\)là dấu nhân
Đặt \(A=\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)
\(B=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2008}+\frac{1}{2009}\)
Ta có :
\(A=\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}\)
\(\Rightarrow A=1+\left(\frac{2007}{2}+1\right)+\left(\frac{2006}{3}+1\right)+...+\left(\frac{2}{2007}+1\right)+\left(\frac{1}{2008}+1\right)\)
\(\Rightarrow A=\frac{2009}{2009}+\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2007}+\frac{2009}{2008}\)
\(\Rightarrow A=\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2007}+\frac{2009}{2008}+\frac{2009}{2009}\)
\(\Rightarrow A=2009.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)\)
\(\Rightarrow A=2009.B\)
Nên : \(\frac{A}{B}=\frac{2009.B}{B}=2009\)
Vậy kết quả biểu thức đã cho là \(2009\)
~ Ủng hộ nhé
\(\frac{\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2008}+\frac{1}{2009}}\)
\(=\frac{\left(\frac{2007}{2}+1\right)+\left(\frac{2006}{3}+1\right)+...+\left(\frac{2}{2007}+1\right)+\left(\frac{1}{2008}+1\right)+1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2008}+\frac{1}{2009}}\)
\(=\frac{\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2007}+\frac{2009}{2008}+\frac{2009}{2009}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2008}+\frac{1}{2009}}\)
\(=\frac{2009.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2008}+\frac{1}{2009}}\)
\(=2009\)
\(\frac{\frac{2008}{1}+\frac{2007}{2}+...+\frac{1}{2008}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}+\frac{1}{2009}}\)
\(=\frac{\frac{1}{1}+\left(1+\frac{2007}{2}\right)+...+\left(1+\frac{1}{2008}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}+\frac{1}{2009}}\)
\(=\frac{\frac{2009}{2}+\frac{2009}{3}+...+\frac{2009}{2009}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2009}}\)
\(=\frac{2009\times\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2009}}\)
\(=2009\)
(2008+2007/2+2006/3+2005/4+....+2/2007+1/2008) / (1/2+1/3+1/4+...+1/2009)
Xét tử
2008+2007/2+2006/3+2005/4+ ... +2/2007+1/2008
=(1+1+1+...+1)+2007/2+2006/3+2005/4+ ... +2/2007+1/2008
= 1+ (2007/2)+1+(2006/3)+1+(2005/4)+1+ ... + (2/2007)+1+(1/2008)+1
=2009/2009+2009/2+2009/3+2009/4+ ... + 2009/2007 + 2009/2008
=2009.(1/2+1/3+1/4+ ... + 1/2007+1/2008+1/2009)
Ta có tử số bằng: 2008+2007/2+2006/3+2005/4+…..+2/2007+1/2008
(Phân tích 2008 thành 2008 con số 1 rồi đưa vào các nhóm)
= (1 + 2007/2) + (1 + 2006/3) + (1 + 2005/4) +... + (1 + 2/2007) + ( 1 + 1/2008) + (1)
= 2009/2 + 2009/3 + 2009//4 + ……. + 2009/2007 + 2009/2008 + 2009/2009
= 2009 x (1/2 + 1/3 + 1/4 + ... + 1/2007 + 1/2008 + 1/2009)
Mẫu số: 1/2 + 1/3 + 1/4 + ... + 1/2007 + 1/2008 + 1/2009
Vậy A = 2009
Tính tỉ số B A , biết: 2008 1 2007 2 ... 3 2006 2 2007 1 2008 2009 1 2008 1 2007 1 ... 4 1 3 1 2 1 = + + + + + = + + + + + + B A
Tính nhanh :a)2/5+4/5 nhân 5/2
b)2008/2009-2009/2008+1/2009+2007/2008