Trả lời

a)\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...\dfrac{2}{99.101}\)

=\(2.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\right)\)

=\(2.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

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

=\(2.\dfrac{100}{101}\)

=\(\dfrac{200}{101}\)

Hình như phần b bạn chép đề sai hay sao đấy

a/\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\)

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

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

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

=\(\dfrac{2}{2}.\dfrac{100}{101}\)

=\(\dfrac{100}{101}\)

Bài b mk k bít làm!Chúc bn hc tốt!

\(B=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\\ B=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\\ B=\dfrac{1}{1}-\dfrac{1}{101}\\ B=\dfrac{101}{101}-\dfrac{1}{101}\\ B=\dfrac{100}{101}\)

A=2.(1/1.3 + 1/3.5 + 1/5.7 +.......+1/99.101)

=2.(1/1 + 1/3 + 1/5 + 1/5 + 1/7 +...+1/99 + 1/101)

=2.(1-1/101)

=2.(101/101-1/101)

=2.100/101

200/101

B=2.(1/1.3+1/3.5+1/3.1+....+1/99.101)

=2.(1/1+1/3+1/3+1/5+1/3+1/7+....+1/99+1/101)

=2.(1/1+1/101)

=2.(101/101+1/101)

=2.102/101

=204/101

C=1/2+1/3+1/3+1/4+....+1/99+1/100

=1/2+1/100

=50/100+1/100

=51/100

A = \(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) + ...........+ \(\dfrac{2}{99.101}\)

A = \(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) +............+ \(\dfrac{1}{99}\) - \(\dfrac{1}{101}\)

A = \(\dfrac{1}{1}\) - \(\dfrac{1}{101}\)

A = \(\dfrac{100}{101}\)

=1/1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101

=1-1/101

=100/101

A=2/1.3+2/3.5+2/5.7+...+2/99.101

=1-1/3+1/3-1/5+1/5-1/7+...1/99-1/101

=1-1/101

=1/100

\(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{99.101}\\ =\dfrac{4}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =2.\left(1-\dfrac{1}{101}\right)\\ =2.\dfrac{100}{101}\\ =\dfrac{200}{101}\)

`4/1.3+4/3.5+4/5.7+...+4/99.101`

`=2(2/1.3+2/3.5+2/5.7+...+2/99.101)`

`=2(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101)`

`=2(1-1/101)`

`=2. 100/101`

`=200/101`

\(M=\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}=\frac{3\left(\frac{1}{5}+\frac{1}{7}-\frac{3}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{3}{4}\) \(\frac{3}{4}\)                                                                                                          \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=2-\frac{2}{101}=\frac{200}{101}\)

\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)

\(B=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)

\(B=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

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

\(B=2.\frac{100}{101}=\frac{200}{101}\)

a, \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ =1-\dfrac{1}{101}=\dfrac{100}{101}\)

b, \(\dfrac{5}{1.3}+\dfrac{5}{3.5}+...+\dfrac{5}{99.101}\)

\(=\dfrac{5}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\right)\)

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

\(=\dfrac{5}{2}\left(1-\dfrac{1}{101}\right)\)

\(=\dfrac{5}{2}.\dfrac{100}{101}=\dfrac{250}{101}\)

Vậy...

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\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

\(=1-\dfrac{1}{101}\)

\(=\dfrac{101}{101}+\dfrac{-1}{101}\)

\(=\dfrac{100}{101}\)

a)

\(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+...+\dfrac{1}{24.25}\)

\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

\(=\dfrac{1}{5}-\dfrac{1}{25}\)

\(=\dfrac{4}{25}\)

b)

\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

\(=1-\dfrac{1}{101}\)

\(=\dfrac{100}{101}\)

a) \(\dfrac{1}{5.6}=\dfrac{1}{5}-\dfrac{1}{6}\)

⇒ \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{24.25}=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}=\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)b) \(\dfrac{2}{1.3}=1-\dfrac{1}{3}\)

tương tự

\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}\)