\(\dfrac{5}{1\cdot3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{201\cdot203}\)

= \(\dfrac{5}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{201\cdot203}\right)\)

= \(\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{201}-\dfrac{1}{203}\right)\)

= \(\dfrac{5}{2}\left(1-\dfrac{1}{203}\right)\)

= \(\dfrac{5}{2}\cdot\dfrac{202}{203}=\dfrac{505}{203}\)

Ta có :

  \(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{5.7}+...+\dfrac{5}{201.203}\)

\(=\dfrac{5}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{201.203}\right)\)

\(=\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...+\dfrac{1}{201}-\dfrac{1}{203}\right)\)

\(=\dfrac{5}{2}\left(1-\dfrac{1}{203}\right)\)

\(=\dfrac{5}{2}.\dfrac{202}{203}\)

\(=\dfrac{505}{203}\)

\(2A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{201\cdot203}\)

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{201}-\frac{1}{203}\)

\(2A=1-\frac{1}{203}\)

\(A=\frac{101}{203}\)

Đáp án là : \(\frac{101}{203}\)

100% là 101 / 203 luôn

a)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\)

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

\(=\frac{1}{2}.\left(1-\frac{1}{7}\right)\)

\(=\frac{1}{2}.\frac{6}{7}\)

\(=\frac{3}{7}\)

b)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009.2011}\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)

\(=\frac{1}{2}.\frac{2010}{2011}\)

\(=\frac{1005}{2011}\)

a)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{99}-\frac{1}{101}\right)\)

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

b) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}=\frac{2}{1.3}.\frac{5}{2}+\frac{2}{3.5}.\frac{5}{2}+\frac{2}{5.7}.\frac{5}{2}+...+\frac{2}{99.101}.\frac{5}{2}\)

                                                                \(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)

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

a)100/101

b)250/101

A)100/101

B)250/101

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

100/101

b.5/1.3+5/3.5+5/5.7+............+5/99.101

5.2/1.3.2+5.2/3.5.2+5.2/5.7.2+........+5.2+99.101.2

5/2(2/1.3+2/3.5+2/5.7+........+2/99.101)

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

5/2(1-1/101)

5/2.100/101

250/101

\(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{201.203}\)

\(=\frac{1}{2}.2.\left(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{201.203}\right)\)

\(=\frac{1}{2}.2.3.\left(\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{201.203}\right)\)

\(=\left(\frac{1}{2}.3\right).2.\left(\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{201.203}\right)\)

\(=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{201.203}\right)\)

Vì muốn chuyển 3/5.7 = 1/5 - 1 /7 thì tử số phải bằng hiệu của mẫu số nên 3/5.7= 3/5.7 chia 2/5.7 = 3/2 . 2/5.7 các phân số khác cũng tương tự như thế

nên ta có 3/5.7 +3/7.9 +...3/201.203 = 3/2. (2/5.7+2/7.9+...+2/201.203)

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

    =1-1/101

    =100/101

b) =(2/1.3+2/3.5+2/5.7+...+2/99.101).2,5

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

    =(1-1/101).2,5

    =100/101.2,5

    =250/101

c) =(2/2.4+2/4.6+2/6.8+...+2/2008-2/2010).2

    =(1/2-1/4+1/4-1/6+1/6-1/8+...+1/2008-1/2010).2

    =(1/2-1/2010).2

    =1004/1005

Ta đặt

  \(A=1\times3+3\times5+...+61\times63\)

\(6A=1\times3\times6+3\times5\times6+....+61\times63\times6\)

\(6A=1\times3\times6+3\times5\times\left(7-1\right)+...+61\times63\times\left(65-59\right)\)

\(6A=1\times3\times6+3\times5\times7-1\times3\times5+...+61\times63\times65-59\times61\times63\)

\(6A=1\times3\times6-1\times3\times5+61\times63\times65\)

\(6A=3+61\times63\times65\)

\(6A=3\times\left(1+61\times21\times65\right)\)

\(2A=83266\)

\(A=83266\div2=41633\)