\(1,\\ =\dfrac{2-1}{1\times2}+\dfrac{3-2}{2\times3}+\dfrac{4-3}{3\times4}+\dfrac{5-4}{4\times5}+.....+\dfrac{99-98}{98\times99}+\dfrac{100-99}{99\times100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{100-1}{100}=\dfrac{99}{100}\)

\(2,=\dfrac{13-11}{11\times13}+\dfrac{15-13}{13\times15}+....+\dfrac{21-19}{19\times21}+\dfrac{23-21}{21\times23}\\ =\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+....+\dfrac{1}{19}-\dfrac{1}{21}+\dfrac{1}{21}-\dfrac{1}{23}\\ =\dfrac{1}{11}-\dfrac{1}{23}\\ =\dfrac{23-11}{11\times23}=\dfrac{12}{253}\)

@seven

a: 1/1*2+1/2*3+...+1/99*100

=1-1/2+1/2-1/3+...+1/99-1/100

=1-1/100

=99/100

b: 2/11*13+2/13*15+...+2/21*23
=1/11-1/13+1/13-1/15+...+1/21-1/23

=1/11-1/23

=12/253

Đặt A = \(\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\)

\(\dfrac{1}{7}A=\dfrac{1}{7}\left(\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\right)\)

\(=\dfrac{5}{2.7}+\dfrac{4}{7.11}+\dfrac{3}{11.14}+\dfrac{1}{14.15}+\dfrac{13}{15.28}\)

\(=\dfrac{7-2}{2.7}+\dfrac{11-7}{7.11}+\dfrac{14-11}{11.14}+\dfrac{15-14}{14.15}+\dfrac{28-15}{15.28}\)

\(=\dfrac{7}{2.7}-\dfrac{2}{2.7}+\dfrac{11}{7.11}-\dfrac{7}{7.11}+\dfrac{14}{11.14}-\dfrac{11}{11.14}+\dfrac{15}{14.15}-\dfrac{14}{14.15}+\dfrac{28}{15.28}-\dfrac{15}{15.28}\)

\(=\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{28}\)

\(=\dfrac{1}{2}-\dfrac{1}{28}=\dfrac{14}{28}-\dfrac{1}{28}=\dfrac{13}{28}\)

\(A=\dfrac{13}{28}\div\dfrac{1}{7}=\dfrac{13}{4}\)

Đặt A = \(\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\)

\(\Rightarrow\dfrac{1}{7}.A=\dfrac{5}{2.7}+\dfrac{4}{7.11}+\dfrac{3}{11.14}+\dfrac{1}{14.15}+\dfrac{13}{15.28}\)

\(\Rightarrow\dfrac{1}{7}.A=\left(\dfrac{1}{2}-\dfrac{1}{7}\right)+\left(\dfrac{1}{7}-\dfrac{1}{11}\right)+\left(\dfrac{1}{11}-\dfrac{1}{14}\right)+\left(\dfrac{1}{14}-\dfrac{1}{15}\right)+\left(\dfrac{1}{15}-\dfrac{1}{28}\right)\)

\(\Rightarrow\dfrac{1}{7}.A=\dfrac{1}{2}-\dfrac{1}{28}=\dfrac{13}{28}\)

\(\Leftrightarrow A=\dfrac{13}{4}\)

Vậy...................

a) 4/9 x ( 3/5 + 8/5 - 2/10 )

= 4/9 x 1

=4/9

b) ( 312 + 325 - 247 ) : 13

= 390 : 13

= 30

a)4/9 x 3/5 + 4/9 x 8/5 - 4/9 x 2/10 

=4/9x(3/5+8/5-2/10)

= 4/9x1

=4/9

b,312 : 13 + 325 : 13 - 247 : 13

=(312+325-247) : 13

=         390 : 3

= 130

`c)(15/(sqrt6+1)+4/(sqrt6-2)-12/(3-sqrt6))*(sqrt6+11)`

`=((15(sqrt6-1))/(6-1)+(4(sqrt6+2))/(6-4)-(12(3+sqrt6))/(9-6))*(sqrt6+11)`

`=(3(sqrt6-1)+2(sqrt6+2)-4(3+sqrt6))*(sqrt6+11)`

`=(3sqrt6-3+2sqrt6+4-12-4sqrt6)*(sqrt6+11)`

`=(sqrt6-11)(sqrt6+11)`

`=6-121=-115`

c) Ta có: \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)

=6-121=-115

=-2/13

\(=\left(\dfrac{2}{9}+\dfrac{7}{9}\right)+\left(\dfrac{3}{7}+\dfrac{4}{7}\right)-\dfrac{2}{13}\)

\(=1+1-\dfrac{2}{13}=2-\dfrac{2}{13}=\dfrac{26}{13}-\dfrac{2}{13}=\dfrac{24}{13}\)

= ( 2/9 + 7/9 ) + ( 3/7 + 4/7 ) - 2/13

= 1 + 1 - 2/13

= 2-2/13

= 26/13 - 2/13 = 23/13

a) = \(\dfrac{12}{17}\)

b) = \(\dfrac{31}{37}\)

a: =9+7=16

b: =11+3/13-2-4/7-5-3/13

=4-4/7

=28/7-4/7=24/7

c: =2/7(5+1/4-3-1/4)=2/7x2=4/7