\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\)

=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)

=\(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+\frac{9-8}{8.9}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)

=\(1-\frac{1}{9}=\frac{8}{9}\)

1/2 + 1/6 + 1/12 + ... + 1/72

= 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9

= 1 - 1/9

= 8/9

Ta có: \(\frac{1}{2}+\frac{1}{-3}+\frac{1}{4}+\frac{1}{6}\)

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

\(=\frac{1}{6}+\frac{1}{6}-\frac{1}{4}\)

\(=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

90:5+110:5=(90+110):5

                 =200:5

                 =40

HT

90:5+110:5                                 90:5+110:5

= (90+110) : 5                            = 18 + 22

= 200         : 5                            =  40

= 40

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{132}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)

\(=\frac{1}{1}-\frac{1}{12}\)

\(=\frac{11}{12}\)

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{132}\)

\(A=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{11x12}\)

\(A=\frac{2-1}{1x2}+\frac{3-2}{2x3}+\frac{4-3}{3x4}+\frac{5-4}{4x5}+...+\frac{12-11}{11x12}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)

\(A=1-\frac{1}{12}=\frac{11}{12}\)

\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+....+\dfrac{1}{110}\)

\(=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+....+\dfrac{1}{10\times11}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{10}-\dfrac{1}{11}\)

\(=1-\dfrac{1}{11}=\dfrac{10}{11}\)

( 168 x 168 - 168 x 58 ) : 110

= [ 168 x ( 168 - 58 ) ] : 110

= 168 x 110 : 110

= 168

168 x 110 : 110 = 168

( 168 x 168 - 168 x 58 ) : 110 = [ 168 x ( 168 - 58 ) ] : 110

                                           = [ 168 x 110 ] : 110

                                           = 18480 : 110

                                           = 168

ko biết

    110+220+330+440+560+670+780+890+500

   = (110+890)+(220+780)+(330+670)+(440+560)+500

   =  1000+1000+1000+1000+500

   =4500

Tính bằng cách thuận tiện nhất

110+220+330+440+560+670+780+890+500

Trả lời :

( 110  + 890 )  ( 220 +  780 ) +  ( 330 + 670 ) +  ( 440 +  560 )  +  560 +  500 = 

 1000 +    1000 +   1000 +    1000 +   560 + 500 =

   4000 + 1060 = 

    5060

^ HT ^

misa

Đặt tên biểu thức là A ta có :

\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+....+\frac{1}{90}+\frac{1}{110}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{10}-\frac{1}{11}\)

\(=\left(\frac{1}{2}-\frac{1}{11}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+.....+\left(\frac{1}{10}-\frac{1}{10}\right)\)

\(=\left(\frac{1}{2}-\frac{1}{11}\right)+0+......+0\)

\(=\frac{11}{22}-\frac{2}{22}=\frac{9}{22}\)

\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{10.11}\)

=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)

=\(\frac{1}{2}-\frac{1}{11}\)

=\(\frac{9}{22}\)

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{110}\)

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}+\frac{1}{10.11}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)

\(=1-\frac{1}{11}\)

\(=\frac{10}{11}\)