1 + 9 + 2 + 8 + 3 + 7 + 4 + 6 + 5 + 5 + 6 + 4 + 7 + 3 + 8 + 2 + 9 + 1

= 10 + 10 + 10 +10 +10 + 10 + 10 + 10 + 10

= 10 x 9

= 90

1+9+2+8+3+7+4+6+5+5+6+4+7+3+8+2+9+1

=10+10+10+10+10

=50

(3-1/4+2/3)-(5+1/3-6/5)-(6-7/4+3/2)=-6/7/15

bằng -523/60

bằng -97/15 

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{8}\)

\(\Rightarrow A=\frac{3}{8}\)

\(A=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{7}-\frac{1}{8}\right)\)

\(A=\frac{1}{2}-\frac{1}{8}\)

\(A=\frac{3}{8}\)

\(A=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)

\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

\(A=\frac{1}{2}-\frac{1}{8}\)

\(A=\frac{3}{8}\)

Đặt S = 1 x 2 + 2 x 3 + ..... + 9 x 10

3S = 1 x 2 x (3-0) + 2 x 3 x (4 - 1) + ....... + 9 x 10 x (11 - 8)

3S = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + ......... + 9 x 10 x 11 - 8 x 9 x 10

3S = ( 1 x 2 x 3 -1 x 2 x 3) + ( 2 x 3 x4 - 2 x 3 x 4) + ..... + (8 x 9  x 10 - 8 x 9 x 10) + 9 x 10 x 11

3S = 9 x 10 x 11 = 990

S = 990 : 3 = 330     

a,  \(2\dfrac{1}{3}+4\dfrac{1}{5}+4\dfrac{1}{3}\)

\(=\dfrac{7}{3}+\dfrac{21}{5}+\dfrac{13}{3}\)

\(=\dfrac{1}{3}\left(7+13\right)+\dfrac{21}{5}\)

\(=\dfrac{20}{3}+\dfrac{21}{5}=\dfrac{100+63}{15}=\dfrac{163}{15}\)

b, \(5\dfrac{3}{4}-4\dfrac{1}{2}.3\dfrac{7}{8}\)

\(=\dfrac{23}{4}-\dfrac{9}{2}.\dfrac{31}{8}\)

\(=\dfrac{23}{4}-\dfrac{279}{16}=\dfrac{92-279}{16}=-\dfrac{187}{16}\)

c, \(1\dfrac{1}{2}.3\dfrac{2}{3}.4\dfrac{3}{4}\)

\(=\dfrac{3}{2}.\dfrac{11}{3}.\dfrac{19}{4}=\dfrac{209}{8}\)

d, \(6\dfrac{4}{5}:2\dfrac{3}{4}:1\dfrac{1}{2}\)

\(=\dfrac{34}{5}:\dfrac{11}{4}:\dfrac{3}{2}\)

\(=\left(\dfrac{34}{5}.\dfrac{4}{11}\right).\dfrac{2}{3}=\dfrac{136}{55}.\dfrac{2}{3}=\dfrac{272}{165}\)

A=1/1*2+1/2*3+...+1/9*10

=1-1/2+1/2-1/3+...+1/9-1/10

=(1-1/10)+(1/2-1/2)+...+(1/9-1/9)

=(10/10-1/10)+0+...+0=9/10

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}=\frac{9}{10}\)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

\(=\frac{1}{2}-\frac{1}{8}\)

\(=\frac{3}{8}\)

Dấu chấm là dấu nhân nhé