2006 . 2005 = 4022030

2004 . 2006 + 2005 = 4022029 

chúc bạn học tốt ^_^

ko hiểu

=\(\frac{2006\cdot2005-1}{2004\cdot2006+2005}\)

=\(\frac{2006\cdot2005-\left(2005\cdot1\right)-1}{2004\cdot2006+2005}\)

=\(\frac{2006\cdot2005-\left(2005-1\right)}{2004\cdot2006+2005}\)

=\(\frac{2006\cdot2005-2004}{2004\cdot2006+2005}\)

=1

mk chỉ giải đc câu d thôi nha ; bn thông cảm

d) \(D=\dfrac{2006.2005-1}{2004.2006+2005}=\dfrac{2006.2005-1}{2004.2006+2006-1}\)

\(D=\dfrac{2006.2005-1}{2006\left(2004+1\right)-1}=\dfrac{2006.2005-1}{2006.2005-1}=1\)

A = (2005 - 1).(2005 +1).(2005² + 1)  = (2005² - 1).(2005² + 1) = 2005⁴ - 1 < 2005⁴

=> A < B

A= 2005⁴-2004.2006.(2005²+1)

=\(2005^4-\left(2005-1\right)\cdot\left(2005+1\right)\cdot\left(2005^2+1\right)\)

=\(2005^4-\left(2005^2-1\right)\cdot\left(2005^2+1\right)\)

=\(2005^4-\left(2005^4-1\right)\)

=1

B=1999.(2000²+2001)-2001.(2000²-1999)

=\(1999\cdot2000^2+1999\cdot2001-2001\cdot2000^2+2001\cdot1999\)

=\(2000^2\left(1999-2001\right)+2\cdot1999\cdot2001\)

=\(2000^2\cdot\left(-2\right)+2\cdot1999\cdot2001\)

=\(2000^2\cdot\left(-2\right)+2\left(2000-1\right)\left(2000+1\right)\)

=\(-2\cdot2000^2+2\left(2000^2-1\right)\)

=\(-2\cdot2000^2+2\cdot2000^2-2\)

=-2

A= 2004.2006.(2005²+1)

=(2005-1)(2005+1)(2005²+1)

=(2005²-1)(2005²+1)

=2005⁴-1<2005⁴

=> A<B

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