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

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

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

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

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

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

`A=2/[1.3]+2/[3.5]+2/[5.7]+.....+2/[99.101]`

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

`A=1-1/101=101-1/101=100/101`

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

A=2/1.3+2/3.5+2/5.7+...+2/99.101
   = 1/1 - 1/3 +1/3 - 1/5 +.... +1/99 - 1/101
   = 1-1/101
   =101/101-1/101
   =100/101   

Ta có :

M= \(\dfrac{3+3-3+\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{4+4-4+\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\)= \(\dfrac{3+3-3}{4+4-4}=\dfrac{3}{4}\)

b) Nhận xét thấy: \(\dfrac{2}{1.3}=1-\dfrac{1}{3};\dfrac{1}{3.5}=\dfrac{1}{3}-\dfrac{1}{5};...\)

Ta có:

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

B= 1- \(\dfrac{1}{101}\)= \(\dfrac{100}{101}\)

Vậy B= \(\dfrac{100}{101}\)

T=4/1 . 4/3 + 4/3 . 4/5 + ... + 4/99 . 4/100

T=4/1 - 4/3 + 4/3 - 4/5 + ... + 4/99 - 4/100

T=4/1 - 4/100

T=99/25

\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)

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

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

\(=\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}.\dfrac{100}{101}=\dfrac{50}{101}\)

2/ = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) +......+\(\dfrac{1}{100.101}\)

= 1-\(\dfrac{1}{2}\) +\(\dfrac{1}{2}\) -\(\dfrac{1}{3}\)+.........+\(\dfrac{1}{100}\)-\(\dfrac{1}{101}\)

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

mk làm vậy thôi nha

thông cảm

=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{4.5}\)=\(1-\dfrac{1}{2}+....+\dfrac{1}{4}-\dfrac{1}{5}\)

=1-\(\dfrac{1}{5}=\dfrac{4}{5}\)

tương tự

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

\(A=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)=\dfrac{1}{2}\left(\dfrac{100}{101}\right)=\dfrac{50}{101}\)

\(A=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)=\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{50}{101}\)

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

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

= 1/2 . 100/101

= 50/101

\(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{19.21}\)

\(=2\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{19.21}\right)\)

\(=2\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\)

\(=2\left(1-\dfrac{1}{21}\right)=2.\dfrac{20}{21}=\dfrac{40}{21}\)