a) \(1+2+3+4+...+n\)

\(=\left(n+1\right)\left[\left(n-1\right):1+1\right]:2\)

\(=\left(n+1\right)\left(n-1+1\right):2\)

\(=n\left(n+1\right):2\)

\(=\dfrac{n\left(n+1\right)}{2}\)

b) \(2+4+6+..+2n\)

\(=\left(2n+2\right)\left[\left(2n-2\right):2+1\right]:2\)

\(=2\left(n+1\right)\left[2\left(n-1\right):2+1\right]:2\)

\(=\left(n+1\right)\left(n-1+1\right)\)

\(=n\left(n+1\right)\)

c) \(1+3+5+...+\left(2n+1\right)\)

\(=\left[\left(2n+1\right)+1\right]\left\{\left[\left(2n-1\right)-1\right]:2+1\right\}:2\)

\(=\left(2n+1+1\right)\left[\left(2n-1-1\right):2+1\right]:2\)

\(=\left(2n+2\right)\left[\left(2n-2\right):2+1\right]:2\)

\(=2\left(n+1\right)\left[2\left(n-1\right):2+1\right]:2\)

\(=\left(n+1\right)\left(n-1+1\right)\)

\(=n\left(n+1\right)\)

d) \(1+4+7+10+...+2005\)

\(=\left(2005+1\right)\left[\left(2005-1\right):3+1\right]:2\)

\(=2006\cdot\left(2004:3+1\right):2\)

\(=2006\cdot\left(668+1\right):2\)

\(=1003\cdot669\)

\(=671007\)

e) \(2+5+8+...+2006\)

\(=\left(2006+2\right)\left[\left(2006-2\right):3+1\right]:2\)

\(=2008\cdot\left(2004:3+1\right):2\)

\(=1004\cdot\left(668+1\right)\)

\(=1004\cdot669\)

\(=671676\)

g) \(1+5+9+...+2001\)

\(=\left(2001+1\right)\left[\left(2001-1\right):4+1\right]:2\)

\(=2002\cdot\left(2000:4+1\right):2\)

\(=1001\cdot\left(500+1\right)\)

\(=1001\cdot501\)

\(=501501\)

đề câu e sai r

a) 1+2+3+4+5+...+n = n(n+1) / 2

b)2+4+6+...+2n = [(2n-2):2+1] . (2n+2)/2 = n . ( 2n+2) /2

C=(-1+3)+(-5+7)+....+(2011-2013)

  = 2+2+2+...+(-2)

  = 1004+(-2)

  = 1002

D= (2-4)+(6-8)+....+(2010-2012)

  =  -2+-2+-2+...1002+...+-2

  = -502+1002

  = 500

G=(1+2-3-4)+(5+6-7-8)+...+(109+110-111-112)+(113+114+115)

  = -4+-4+-4+...+-4+342

  =-112+342

  = 230 

Bài 7:

7.1: I là trung điểm của AB

=>\(AB=2\cdot IA=4\left(cm\right)\)

7.2:

C nằm giữa A và B

=>AC+CB=AB

=>CB=10-8=2(cm)

C là trung điểm của NB

=>NC=CB=2cm

C là trung điểm của NB

=>\(NB=2\cdot NC=2\cdot2=4\left(cm\right)\)

Bài 6:

a: \(\dfrac{4}{5}=\dfrac{4\cdot6}{5\cdot6}=\dfrac{24}{30}\)

\(\dfrac{8}{15}=\dfrac{8\cdot2}{15\cdot2}=\dfrac{16}{30}\)

\(-\dfrac{3}{2}=\dfrac{-3\cdot15}{2\cdot15}=-\dfrac{45}{30}\)

b: \(2=\dfrac{2\cdot45}{45}=\dfrac{90}{45}\)

\(\dfrac{-10}{5}=\dfrac{-10\cdot9}{5\cdot9}=\dfrac{-90}{45}\)

\(\dfrac{7}{-9}=\dfrac{-7}{9}=\dfrac{-7\cdot5}{9\cdot5}=\dfrac{-35}{45}\)

c: \(\dfrac{3}{-2}=\dfrac{-3}{2}=\dfrac{-3\cdot6}{2\cdot6}=\dfrac{-18}{12}\)

\(\dfrac{5}{-6}=\dfrac{-5}{6}=\dfrac{-5\cdot2}{6\cdot2}=\dfrac{-10}{12}\)

\(\dfrac{-6}{4}=\dfrac{-6\cdot3}{4\cdot3}=\dfrac{-18}{12}\)

d: \(-\dfrac{1}{2}=\dfrac{-1\cdot15}{2\cdot15}=\dfrac{-15}{30}\)

\(\dfrac{4}{3}=\dfrac{4\cdot10}{3\cdot10}=\dfrac{40}{30}\)

\(\dfrac{6}{-5}=\dfrac{-6}{5}=\dfrac{-6\cdot6}{5\cdot6}=\dfrac{-36}{30}\)

bài 5:

a: \(\dfrac{3}{4}=\dfrac{9}{12};\dfrac{-3}{12}=\dfrac{-3}{12};\dfrac{-2}{3}=-\dfrac{8}{12};\dfrac{-1}{-6}=\dfrac{1}{6}=\dfrac{2}{12}\)

mà -8<-3<2<9

nên \(-\dfrac{8}{12}< -\dfrac{3}{12}< \dfrac{2}{12}< \dfrac{9}{12}\)

=>\(\dfrac{-2}{3}< \dfrac{-3}{12}< \dfrac{-1}{-6}< \dfrac{3}{4}\)

b: Ta có: \(\dfrac{-7}{9}=\dfrac{-28}{36};\dfrac{-1}{3}=\dfrac{-12}{36};-1=-\dfrac{36}{36}\)

mà -36<-28<-12

nên \(-1< -\dfrac{28}{36}< -\dfrac{12}{36}\)

=>\(-1< \dfrac{-7}{9}< -\dfrac{1}{3}< 0\)

\(\dfrac{5}{12}=\dfrac{15}{36};\dfrac{-1}{-4}=\dfrac{1}{4}=\dfrac{9}{36}\)

mà 9<15

nên \(0< \dfrac{1}{4}< \dfrac{5}{12}\)

=>\(-1< -\dfrac{7}{9}< -\dfrac{1}{3}< 0< \dfrac{1}{4}< \dfrac{5}{12}\)

c: \(\dfrac{-1}{-2};0;\dfrac{3}{10};1;\dfrac{-2}{-5};\dfrac{3}{-4}\)

\(-\dfrac{3}{4}< 0\)

\(\dfrac{-1}{-2}=\dfrac{1}{2}=\dfrac{5}{10};\dfrac{3}{10}=\dfrac{3}{10};1=\dfrac{10}{10};\dfrac{-2}{-5}=\dfrac{4}{10}\)

mà 3<4<5<10

nên \(\dfrac{3}{10}< \dfrac{4}{10}< \dfrac{5}{10}< \dfrac{10}{10}\)

=>\(0< \dfrac{3}{10}< \dfrac{-2}{-5}< \dfrac{-1}{-2}< 1\)

=>\(-\dfrac{3}{4}< 0< \dfrac{3}{10}< \dfrac{-2}{-5}< \dfrac{-1}{-2}< 1\)

d: \(-\dfrac{37}{150}=\dfrac{-37}{150};\dfrac{17}{-50}=\dfrac{-17}{50}=\dfrac{-51}{150}\)

\(\dfrac{23}{-25}=\dfrac{-23}{25}=\dfrac{-138}{150};\dfrac{-7}{10}=\dfrac{-105}{150};\dfrac{-2}{5}=-\dfrac{60}{150}\)

mà -138<-105<-60<-51<-37

nên \(-\dfrac{138}{150}< -\dfrac{105}{150}< -\dfrac{60}{150}< -\dfrac{51}{150}< -\dfrac{37}{150}\)

=>\(\dfrac{23}{-25}< \dfrac{-7}{10}< \dfrac{-2}{5}< \dfrac{-17}{50}< \dfrac{37}{-150}\)