A=1+1/2+1/3+1/6+(1/12+1/15+1/20+1/30)+1/35

=71/35+7/30=95/42

\(B=\frac{6}{1\cdot3}+\frac{6}{3\cdot5}+\cdot\cdot\cdot+\frac{6}{97\cdot99}\)

\(\Rightarrow B=3\cdot\left(\frac{2}{1\cdot3}+\cdot\cdot\cdot+\frac{2}{97\cdot99}\right)\)

\(\Rightarrow B=3\cdot\left(1-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{97}-\frac{1}{99}\right)\)

\(\Rightarrow B=3\cdot\left(1-\frac{1}{99}\right)\)

\(\Rightarrow B=3\cdot\frac{98}{99}\)

\(\Rightarrow B=\frac{98}{33}\)

\(A=\frac{1}{2}+\frac{1}{6}+\cdot\cdot\cdot+\frac{1}{42}\)

\(\Rightarrow A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{6\cdot7}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{6}-\frac{1}{7}\)

\(\Rightarrow A=1-\frac{1}{7}\)

\(\Rightarrow A=\frac{6}{7}\)

C=1/2+1/4+1/8+1/16+1/32+1/64+1/128

C=1-1/2+1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32+1/32-1/64+1/64-1/128

C=1-1/128

C=127/128

D=1/2+1/6+1/12+1/20+1/30+1/42

D=1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7

D=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7

D=1/1-1/7

D=6/7

\(a,\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

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

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

\(=1-\frac{1}{7}\)

\(=\frac{6}{7}\)

\(b,\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)

Ta có :

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

\(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)

\(\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\)

\(\frac{1}{16}=\frac{1}{8}-\frac{1}{16}\)

\(\frac{1}{32}=\frac{1}{16}-\frac{1}{32}\)

Thay vào ta có :

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}\)

\(=1-\frac{1}{32}\)

\(=\frac{31}{32}\)

\(c,\)\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)

Ta có :

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

\(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)

...................

\(\frac{1}{256}=\frac{1}{128}-\frac{1}{256}\)

Thay vào ta có :

\(=\)\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{128}-\frac{1}{256}\)

\(=1-\frac{1}{256}\)

\(=\frac{255}{256}\)

133/99

quy đòng r tính nha ra \(\dfrac{199}{33}\)

`a,`

`31/23-(7/32+8/23)`

`=31/23-7/32-8/23`

`=(31/23-8/23)-7/32`

`=1-7/32=25/32`

`b,`

`38/45-(8/45-17/51-3/11)`

`=38/45-8/45+17/51+3/11`

`= (38/45-8/45)+17/51+3/11`

`=2/3+17/51+3/11`

`=1+3/11=14/11`

`c,`

`(1/3+12/67+13/41)-(79/67-28/41)`

`= 1/3+12/67+13/41-79/67+28/41`

`= 1/3+(12/67-79/67)+(13/41+28/41)`

`= 1/3+(-1)+1=1/3+(-1+1)=1/3+0=1/3`

`d,`

`1/5+(-1/6)+1/7+(-1/8)+1/9+1/8+(-1/7)+1/6+(-1/5)`

`= (1/5+ -1/5)+(-1/6+1/6)+(1/7+ -1/7)+(-1/8 +1/8)+1/9`

`= 0+0+0+0+1/9=1/9 .`

A = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72

   = 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- 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

   = 1 - 1/9 = 8/9

Câu B, C dấu * là nhân hay công vậy?