\(M:N=\frac{\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}\)

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 
\(\Rightarrow M:N=\frac{2009.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}=2009\)

a) m x 2 + n x 2 + p x 2

= ( m + b + p ) x 2

= ( 2006 + 2007 + 2008 ) x 2 

= 6021 x 2

= 12042

b) m x 2 + n x 2 + p x 2

= ( m + n + p ) x 2

= 2009 x 2 

= 4018

Đúng 100%

Đúng 100%

Đúng 100%

a/ 

\(m\times2+n\times2+p\times2\)

\(=2006\times2+2007\times2+2008\times2\)

\(=2\left(2006+2007+2008\right)\)

\(=2\times6021\)

\(=12042\)

b/ 

\(m\times2+n\times2+p\times2\)

\(=2\left(m+n+p\right)\)

\(=2\times2009\)

\(=4018\)

m x 2 + n x 2 + p x 2 biết:

a) m = 2006, n = 2007, p = 2008

2006 x 2 + 2007 x 2 + 2008 x 2

= (2006 + 2007 + 2008) x 2

=6021 x 2 = 12042

b) m + n + p = 2009

ta có: 

m x 2 + n x 2 + p x 2

= ( m + n + p) x 2

mà m + n + p = 2009

suy ra : 

m x 2 + n x 2 + p x 2 

= 2009 x 2 =4018

a) m x 2 + n x 2 + p x 2
 = (m + n + p) x 2
 = (2006 + 2007 + 2008) x 2
 = 6021 x 2
 = 12042
b) m x 2 + n x 2 + p x 2
 = (m + n + p) x 2
 = 2009 x 2
 = 4018

a)m x 2 + n x 2 + p x 2 = 2006 x2 +2007 x 2 +2008 x2 = 12042

b)m x 2 + nx2 + px2 = (m+n+p) x2 = 2009x2 = 4018

Ý 1:

Thay m=2006, n= 2007, p=2008 vào biểu thức ta được:

m x 2 + n x 2 + p x 2

= 2006 x 2 + 2007 x 2 + 2008 x 2

= 4012 + 4014 + 4016

= 12042

___

Ý 2:

m x 2 + n x 2 + p x 2

= (m+n+p) x 2

= 2009 x 2

= 4018

a) chắc bạn cũng biết 

b) \(\frac{2005.2007-1}{2004+2005.2006}=\frac{2005.\left(2006+1\right)-1}{2004+2005.2006}=\frac{2005.2006+2005.1-1}{2004+2005.2006}=\frac{2005.2006+2004}{2004+2005.2006}=1\)