\(A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(A=\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-\frac{1}{6.5}-\frac{1}{5.4}-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1}\)

\(-A=\left(\frac{1}{10.9}+\frac{1}{9.8}+\frac{1}{8.7}+\frac{1}{7.6}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

\(-A=\frac{1}{10}-\frac{1}{9}+\frac{1}{9}-\frac{1}{8}+\frac{1}{8}-\frac{1}{7}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\)

\(-A=\frac{1}{10}-1=\frac{-9}{10}\Rightarrow A=\frac{9}{10}\)

\(A=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(=\frac{1}{90}-\left(\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)

\(=\frac{1}{90}-\left(\frac{1}{8.9}+\frac{1}{7.8}+\frac{1}{6.7}+\frac{1}{5.6}+\frac{1}{4.5}+\frac{1}{3.4}+\frac{1}{2.3}+\frac{1}{1.2}\right)\)

\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\)

\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)=\frac{1}{90}-\frac{8}{9}=-\frac{79}{90}\)

Vậy A=-79/90

A=1/1.2-1/2.3-1/3.4-1/4.5-1/5.6-1/6.7-1/7.8-1/8.9-1/9.10(mình làm ngược lại nha)

A=1/1-1/2-1/2-1/3-1/3-1/4-1/4-1/5-1/5-1/6-1/6-1/7-1/7-1/8-1/8-1/9

A=1/1-1/9

A=8/9

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

\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{10\cdot11}\)

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

\(=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)

= 1/2.3 +1/3.4+1/4.5+...+1/10.11

=1/2-1/3+1/3-1/4+1/4-1/5+...+1/10-1/11

=1/2-1/11

=9/22

~~~~~~~~~~~~ chúc bạn học thật giỏi~~~~~~~~~~~~~

Đặt A=1/6 + 1/12 + 1/20 +.........+1/110

A= 1/2x3 +1/3x4 +1/4x5 +......+ 1/10x11

A= 1/2-1/3 +1/3-1/4 +...........+ 1/10-1/11

A=(1/2-1/11) + (1/3 - 1/3) +.......+ (1/10 - 1/10)

A=11/22 -2/22 + 0 +...........+ 0

A=9/22

Vậy A= 9/22

mình biến đởi phần trong |......| rồi bạn thay vào nha

1/30 + 1/42 + 1/56 + 1/72 +1/ 90 + 1/110 + 1/132

=1/5.6  +  1/6.7  +  1/7.8  +  1/8.9  + 1/9.10  +1/ 10.11

=1/5 -1/6 +1/6 - 1/7 +......+1/10 - 1/11

=1/5 - 1/11=11/55 - 5/55 =6/ 55

thay vào |....|=> |6/55 - x | = 2/3 => mở ra 2 trường hợp mà tính nha

chúc hok tốt

=>(1/5.6+1/6.7+1/7.8+1/9.10+1/10.11+1/11.12)-x=2/3

=>(1/5-1/+1/6-1/7+...+1/11-1/12)-x=2/3

=>(1/5-1/12)-x=2/3

=>7/60-x=2/3

=>x=7/60-2/3

=>x=-11/20

| 1/5.6 + 1/6.7+1/7.8+1/8.9+1/9.10+1/10.11+1/11.12 - X| = 2/3

| 1/5.1/6+1/6.1/7+1/7.1/8+1/8.1/9+1/9.1/10+1/10.1/11+1/11.1/12 - X| = 2/3

| 1/5.1/12  - X| = 2/3

| 1/60 - X | =2/3

-X=2/3-1/60

-X=13/20     =>> VÌ X NẰM TRONG GIÁ TRỊ TUYỆT ĐỐI NÊN X= 13/20 VÀ X= -13/20

( CÓ THỂ LÀ SỄ SAI BỞI MÌNH TỚ LƯỢC BỎ 1 SỐ BƯỚC CHO BỚT DÀI DÒNG)

A = \(\frac{-79}{90}\)

B = \(\frac{8}{9}\)

cách giải sao chỉ mình với

1/2+6+12+20+30+42+56+72+90

=1/90

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)

\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}\)

\(=\frac{9}{10}\)

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)

\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)

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

\(=1-\frac{1}{10}\)

\(=\frac{9}{10}\)

_\(\frac{1}{90}\)_ (\(\frac{1}{72}\)+\(\frac{1}{56}\)+\(\frac{1}{42}\)+\(\frac{1}{30}\)+\(\frac{1}{20}\)+\(\frac{1}{12}\)+\(\frac{1}{6}\)+\(\frac{1}{2}\))

Bạn tự làm tiếp nha 

=10/11 đó

Kq:\(\frac{-79}{90}\)   => ủng hộ nhá

\(\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(=\frac{1}{90}-\left(\frac{1}{72}+\frac{1}{56}+\frac{1}{42}+\frac{1}{30}+\frac{1}{20}+\frac{1}{12}+\frac{1}{6}+\frac{1}{2}\right)\)

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

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

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

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

Vậy tổng=1/90+8/9=9/10