Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lời giải:
a)
\(2006.2005^{2003}> 2005.2005^{2003}=2005^{1+2003}=2005^{2004}\)
Vậy \(2006.2005^{2003}> 2005^{2004}\)
b)
\(2005^{2004}+2005^{2003}=2005^{2003}(2005+1)=2005^{2003}.2006< 2006^{2003}.2006\)
hay \(2005^{2004}+2005^{2003}< 2006^{2004}\)
c) Thiếu đề
d)
\(72^{27}-72^{26}=72^{26}(72-1)=71.72^{26}\)
\(72^{28}-72^{27}=72^{27}(72-1)=71.72^{27}> 71.72^{26}\)
\(\Rightarrow 72^{28}-72^{27}> 72^{27}-72^{26}\)
Giải
Ta gọi T = (1^2+2^2+...+2005^2)-(1.3+2.4+3.5+...+2004.2006)
Đặt A = 1^2+2^2+3^2+...+2005^2
=> A = 1.1 + 2.2 +3.3 +...+ 2005.2005
=> A = 1.(2-1) + 2.(3-1) + 3.(4-1) +...+ 2005.(2006-1)
==> A = 1.2-1.1 + 2.3-1.2 + 3.4-1.3+...+2005.2006-1.2005
=> A = (1.2+2.3+3.4+...+2005.2006)-(1+2+3+...+2005)
Xét 1.2 +2.3+3.4+...+2005.2006
= 1/3.(1.2.3+2.3.3+...+2005.2006.3)
=1/3.[1.2.(3-0)+2.3.(4-1)+...+2005.2006.(2007-2004)]
=1/3.(1.2.3+2.3.4-1.2.3+...+2005.2006.2007-2004.2005.2006)
= 1/3 . 2005.2006.2007
= 2005.2006.2007/3 = 2690738070
Vậy A= 2690738070 - (1+3+5+...+2005)
=> A= 2690738070- [(2005-1):2+1].(2005+1)/2
=> A = 2690738070 - 1006009
=> A = 2689732061
Đắt B = 1.3+2.4+3.5+4.6+...+2003.2005 +2004.2006
=> B= (1.3+3.5+...+2003.2005)+(2.4+4.6+...+2004.2006)
=> 6B = (1.3.6+3.5.6+...+2003.2005.6)+(2.4.6+4.6.6+...+2004.2006.6)
=> 6B = [1.3.(5+1)+3.5.(7-1)+...+2003.2005.(2007-2001)] + [2.4.(6-0)+4.6.(8-2)+...+2004.2006.(2008-2002)]
=> 6B = (1.3.5+1.3.1+3.5.7-1.3.5+...+2003.2005.2007-2001.2003.2005)+(2.4.6+4.6.8-2.4.6+...+2004.2006.2008-2002.2004.2006)
=> 6B = 1.3.1+2003.2005.2007 + 2004.2006.2008
=> 6B = 16132350300
=> B = 16132350300/6 = 2688725050
Vì T = A - B = 2689732061-2688725050
=> T = 1007011
2005.2005-2004.2006
=(2004+1).2005-2004.2006
=2004.2005+2005-2004.2006
=(2004.2005+2004)+1-2004.2006
=2004.(2005+1)+1-2004.2006
=2004.2006+1-2004.2006=1
Ta có:2005*2005-2004*2006=2005*2005-2004*(2005+1)=2005*2005-2004*2005+2004=(2005*2005-2004*2005)+2004
=(2005*(2005-2004))+2004=2005*1+2004=2005+2004=4009
Tổng dãy số trên là:
( 2006 + 1 ) . [ ( 2006 - 1 ) : 1 +1 ] : 2 = 2013021
Đ/s: 2013021
\(1+2+3+...+2005+2006=\left[\left(2006-1\right):1+1\right].\left(2006+1\right):2.\)
\(=2006.2007:2\)
\(=2013021\)
\(5+10+...+2000+2005=\left[\left(2005-5\right):5+1\right].\left(2005+5\right):2.\)
\(=401.2010:2\)
\(=403005\)
\(140+136+....+64+60=\left[\left(140-60\right):16+1\right].\left(140+60\right):2.\)
\(=6.200:2\)
\(=600\)
\(1)5-(1997-2005)+1997\)
\(=5-1997+2005+1997\)
\(=(-1997+1997)+(5+2005)\)
\(=0+2010\)
\(=2010\)
\(2)4567+(1234-4567)-4\)
\(=4567+1234-4567-4\)
\(=(4567-4567)+(1234-4)\)
\(=0+1230\)
\(=1230\)
\(-1-2-3-...-2005-2006-2007\)
\(=-\left(1+2+3+...+2005+2006+2007\right)\)
\(=-\frac{\left(2007+1\right).2007}{2}\)
\(=-2015028\)
\(\frac{2006.2005-1}{2004.2006+2005}\)
\(\Leftrightarrow\)\(\frac{2006.\left(2004+1\right)-1}{2004.2006+2005}\)
\(\Leftrightarrow\frac{2006.2004+2016-1}{2004.2006+2005}\)
\(\Leftrightarrow\frac{2006.2004+2005}{2004.2006+2005}\)
\(=1\)
bằng 1 đó k nha