99x101x103:6

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

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

\(\Rightarrow2A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}...+\frac{2}{99.101}\)

\(\Rightarrow2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)

\(\Rightarrow2A=\frac{1}{3}-\frac{1}{101}\)

\(\Rightarrow2A=\frac{101}{303}-\frac{3}{303}\)

\(\Rightarrow2A=\frac{98}{303}\)

\(\Rightarrow A=\frac{98}{303}:2=\frac{98}{303.2}=\frac{98}{606}=\frac{49}{303}\)

lên 820 điểm hỏi đáp nha

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

\(A=1\times3+3\times5+5\times7+...+99\times101\)

\(=1\left(1+2\right)+3\left(3+2\right)+5\left(5+2\right)+...+99\left(99+2\right)\)

\(=\left(1^2+3^2+5^2+...+99^2\right)+2\left(1+3+5+...+99\right)\)

Ta có:

\(1^2+2^2+3^2+...+n^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}\)

⇒ \(A=\left(1^2+2^2+3^2+...+100^2\right)-2^2\left(1^2+2^2+3^2+...+50^2\right)+2\left(1+3+5+...+99\right)\)

\(=\dfrac{100.101.201}{6}+\dfrac{4.50.51.101}{6}+\dfrac{\left(99+1\right).\left[\left(99-1\right):2+1\right]}{2}\)

\(=338350-171700+5000\)

\(=166650+5000=171650\)

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

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

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

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{101}\right)\)

\(=\frac{1}{2}.\frac{98}{303}\)

\(=\frac{49}{303}\)

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

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

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

\(=\frac{1}{2}.\frac{98}{101}=\frac{49}{101}\)

Hình như Nguyễn Hữu Thế trừ sai.

1/3 - 1/101 = 98/303