a) A = 2 + 4 + 6 + 8 + ... + 1000

Ta có : A = 2 + 4 + 6 + 8 + ... + 1000 ( có 500 số )

               = (1000 + 2) . 500 : 2 = 250500

c) \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}+\frac{2}{99.101}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\)

\(=1-\frac{1}{101}=\frac{100}{101}\)

Ta có:

1/30+1/42+1/56+1/72+1/90+1/110+1/132 -x = 2/3

=>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+...+1/11-1/12 - x = 2/3

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

=>7/60 - x = 2/3

=> x = 7/60 - 2/3

=> x = -11/20

Ta có:

1/30+1/42+1/56+1/72+1/90+1/110+1/132 -x = 2/3

=>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+...+1/11-1/12 - x = 2/3

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

=>7/60 - x = 2/3

=> x = 7/60 - 2/3

=> x = -11/20

\(\Rightarrow\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+\dfrac{1}{7\times8}+...+\dfrac{1}{11\times12}=\dfrac{21}{x}\\ \Rightarrow\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{12}=\dfrac{21}{x}\\ \Rightarrow\dfrac{1}{5}-\dfrac{1}{12}=\dfrac{21}{x}\\ \Rightarrow\dfrac{21}{x}=\dfrac{7}{60}\Rightarrow x=\dfrac{21\cdot60}{7}=180\)

\(\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}=\dfrac{21}{x}\)

\(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}=\dfrac{21}{x}\)

\(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}=\dfrac{21}{x}\)

\(\dfrac{1}{5}-\dfrac{1}{12}=\dfrac{21}{x}\)

Còn lại bạn tự tính

b: =8,12(6+8-4)=8,12x10=81,2

a: \(A=\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{10\cdot11}\)

\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{10}-\dfrac{1}{11}\)

=1/4-1/11=7/44

c: =>0,2a+0,4a=12

=>0,6a=12

hay a=20

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)

Giải:

\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-...-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)

\(\Leftrightarrow\dfrac{1}{1.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}-...-\dfrac{1}{9.10}-\dfrac{1}{10.11}=x-\dfrac{5}{13}\)

\(\Leftrightarrow\dfrac{-1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\right)=x-\dfrac{5}{13}\)

\(\Leftrightarrow\dfrac{-1}{2}\left(\dfrac{1}{1}-\dfrac{1}{11}\right)=x-\dfrac{5}{13}\)

\(\Leftrightarrow\dfrac{-1}{2}.\dfrac{10}{11}=x-\dfrac{5}{13}\)

\(\Leftrightarrow\dfrac{-5}{11}=x-\dfrac{5}{13}\)

\(\Leftrightarrow\dfrac{-5}{11}+\dfrac{5}{13}=x\)

\(\Leftrightarrow x=\dfrac{-10}{143}\)

Vậy ...

Giải đẳng cấp nhỉ @Hắc Hường

Ta có:

\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-...-\dfrac{1}{110}=x-\dfrac{5}{13}\\ -\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-...-\dfrac{1}{110}=x-\dfrac{5}{13}-\dfrac{1}{3}\\ -\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-...-\dfrac{1}{110}=x-\dfrac{28}{39}\\ -\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{110}\right)=x-\dfrac{28}{39}\\ -\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{10.11}\right)=x-\dfrac{28}{39}\\ -\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{10}-\dfrac{1}{11}\right)=x-\dfrac{28}{39}\\ -\left(\dfrac{1}{3}-\dfrac{1}{11}\right)=x-\dfrac{28}{39}\\ -\dfrac{8}{33}=x-\dfrac{28}{39}\\ x=-\dfrac{8}{33}+\dfrac{28}{39}\\ x=\dfrac{68}{143}\)

Vậy \(x=\dfrac{68}{143}\)