A=\(\dfrac{113}{140}\)

A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}-\dfrac{1}{20}\)

\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}-\dfrac{1}{20}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}-\dfrac{1}{20}\)

\(=1-\dfrac{1}{7}-\dfrac{1}{20}=1-\dfrac{27}{140}=\dfrac{113}{140}\)

Giải:

A=1/2+1/6+1/12+1/30+1/42

A=1/1.2+1/2.3+1/3.4+1/5.6+1/6.7

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

A=1/1-1/4+1/5-1/7

A=113/140

Chúc bạn học tốt!

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

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

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

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

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

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

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

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

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

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

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

\(D=\frac{1}{2}+\frac{1}{4}+\cdot\cdot\cdot+\frac{1}{128}\)

\(\Rightarrow2D=1+\frac{1}{2}+\cdot\cdot\cdot+\frac{1}{64}\)

\(\Rightarrow2D-D=\left(1+\cdot\cdot\cdot+\frac{1}{64}\right)-\left(\frac{1}{2}+\cdot\cdot\cdot+\frac{1}{128}\right)\)

\(\Rightarrow D=1-\frac{1}{128}\)

\(\Rightarrow D=\frac{127}{128}\)

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

\(=\dfrac{1}{182}\)

1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 = 1/182

1/182

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

Mình chỉnh lại đề B nha:

\(B=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{9999}\)

\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{101}\right)\)

\(=\frac{1}{2}.\frac{100}{101}=\frac{50}{101}\)

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

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

\(A=\frac{99}{100}\)

A=1/1.1/2+1/2.1/3+1/3.1/4+....+1/99.1/100=1-1/100=99/100 ( vì 1/n.1/(n+1)=1/n-1/(n+1) )

(sửa lại phân số thứ 2 của B thành 1/15 mới đúng)

B=1/1.1/3+1/3.1/5+1/5.1/7+....+1/99.1/101=1/2(2/1.2/3+2/3.2/5+2/5.2/7+...+2/99.2/101)=1/2.(1-1/101)=50/101 ( vì 1/n.1/(n+2)=2/n-2/(n+2) )

Mình nghĩ cái phân số đầu tiên bài 2 phải là 11/12 thì: 11/12=33/36; 10/8=45/36; 21/24=14/16

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+1/9-1/9+1/10-(1-1/3+1/3-3/5+3/5-4/7+5/9-5/9+6/11-6/11-7/13)=1+1/10-1+7/13=83/130

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

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