lấy máy tính ra mà tính

a, Theo bài ra ta có:

\(M=\dfrac{2007}{1}+1+\dfrac{2006}{2}+1+.......+\dfrac{2}{2006}+1+\dfrac{1}{2007}+1-2007\)

( Ta thêm 1 vào mỗi một số hạng trong M nên phải bớt đi 2017 vì có 2017 số hạng ) ;'

\(=>M=2008+\dfrac{2008}{2}+\dfrac{2008}{3}+......+\dfrac{2008}{2007}+\dfrac{2008}{2007}-2007\)

\(=>M=\dfrac{2008}{2}+\dfrac{2008}{3}+\dfrac{2008}{4}+.....+\dfrac{2008}{2006}+\dfrac{2008}{2007}+1\)

Ta thấy xuất hiện 2008 chung nên đặt ra ngoài ta có:

\(=>M=2008\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+....+\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}\right)\)

\(=>M:N=2008\)

Câu b đợi 1 chút nha.......

b, \(M=\dfrac{1}{11.13}+\dfrac{1}{13.15}+...+\dfrac{1}{31.33}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{11.13}+\dfrac{2}{13.15}+...+\dfrac{2}{31.33}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{31}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{11}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{33}\)

\(N=\dfrac{12}{11.13.15}+\dfrac{12}{13.15.17}+...+\dfrac{12}{31.33.35}\)

\(=3\left(\dfrac{4}{11.13.15}+\dfrac{4}{13.15.17}+...+\dfrac{4}{31.33.35}\right)\)

\(=3\left(\dfrac{1}{11.13}-\dfrac{1}{13.15}+\dfrac{1}{13.15}-\dfrac{1}{15.17}+...+\dfrac{1}{31.33}-\dfrac{1}{33.35}\right)\)

\(=3\left(\dfrac{1}{11.13}-\dfrac{1}{33.35}\right)\)

\(=\dfrac{92}{5005}\)

\(\Rightarrow M:N=\dfrac{1}{33}:\dfrac{92}{5005}=\dfrac{455}{276}\)

Vậy...

3⁴ . 71 - 3⁴ . 2⁹ 

3⁴ . ( 71 - 2⁹ )

3⁴ . -441 

= -35721

(3³ . 5² - 2⁴ - 16 ) . 13 

= ( 27 . 25- 16 - 16 ) . 13 

= ( 675 - 16 -16 ) . 13 

=  643. 13 

= 8359

35.237 +33.35 

=35 . ( \(237+33\) ) 

= 35 . 270 

= 9450 

  2³ . 4² + 2³ . 84 - 40 

=2³ . ( 16 + 84 ) - 40 

= 8 . 100 - 40 

= 800 - 40 

= 760 

Trả lời:

\(\frac{34}{34}>\frac{33}{35}\)

\(\frac{1991}{1999}< \frac{1995}{1995}\)

Học tốt

a, Ta có: 

\(M=\frac{1}{11.13}+\frac{1}{13.15}+...+\frac{1}{33.35}\)

\(=\frac{1}{2}\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{33}-\frac{1}{35}\right)\)

\(=\frac{1}{2}\left(\frac{1}{11}-\frac{1}{35}\right)=\frac{1}{2}\cdot\frac{24}{385}=\frac{12}{385}\)

\(N=\frac{12}{11.13.15}+\frac{12}{13.15.17}+...+\frac{12}{31.33.35}\)

\(=3\left(\frac{1}{11.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.17}+...+\frac{1}{31.33}-\frac{1}{33.35}\right)\)

\(=3\left(\frac{1}{11.13}-\frac{1}{33.35}\right)=3\cdot\frac{92}{15015}=\frac{92}{5005}\)

\(\Rightarrow M:N=\frac{12}{385}:\frac{92}{5005}=\frac{39}{23}\)

b, \(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}-\left(1+\frac{1}{2}+...+\frac{1}{1006}\right)\)

\(=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2013}\)

\(\Rightarrow S-P-1=\left(\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2013}\right)-\left(\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2013}\right)-1=0-1=-1\)

cam on nha