\(\frac{1}{3}+\frac{13}{15}+\frac{33}{35}+...+\frac{9997}{9999}=1-\frac{2}{3}+1-\frac{2}{15}+1-\frac{2}{35}+...+1-\frac{2}{9999}\)

\(=\left(1+1+1+...+1\right)-\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{9999}\right)\)

\(=50-\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)

\(=50-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(=50-\left(1-\frac{1}{101}\right)=50-\frac{100}{101}=\frac{4950}{101}\)

thank you bạn nhé mình sẽ k cho bạn

nhưng mà sao bạn biết là có 50 số 1

\(A=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+...+\frac{1}{9999}\)

\(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{99.101}\)

\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\right)\)

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

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

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

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

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

2A=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)

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

2A=\(\frac{98}{303}\)

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

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

Chúc bạn học tốt!

Ta có:\(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\)

\(=\frac{1}{3}+\left(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}\right)+\left(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\right)\)\(< \frac{1}{3}+\left(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}\right)+\left(\frac{1}{45}+\frac{1}{45}+\frac{1}{45}\right)\)\(=\frac{1}{3}+\frac{1}{10}+\frac{1}{15}=\frac{1}{2}\)

Vậy ............

Ta có: 1/3 + 1/31 + 1/35 + 1/37 + 1/47 + 1/53 + 1/61 < 1/3 + 3/31 + 3/47 < 1/3 + 3/30 + 3/45

= 1/3 + 1/10 + 1/15 = 1/3 + (1/30) * (3+2) = 1/3 + (1/0) * 5 = 1/3 + 1/6

= (1/6) * (2+1) = (1/6) * 3 = 1/2.

=> 1/3 + 1/31 + 1/35 + 1/37 + 1/47 + 1/53 + 1/61 < 1/2.

Ủng hộ mk nha mina^^

cac ban cho mk biet tai sao lai co phan so \(\frac{1}{30};\frac{1}{45}\)vay ???

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

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

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

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

\(\Rightarrow A=\frac{1}{2}.\frac{88}{303}\)

\(\Rightarrow A=\frac{44}{303}\)

\(A=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{99\times101}\)

\(\Rightarrow2A=\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{99\times101}\)

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

\(=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)

=> A = 98/203 : 2 = 49/303

\(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+...+\frac{1}{99\cdot101}\)

\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\)

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

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

A=1/1*3+1/3*5+1/5*7+.....+1/99*101

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

A=1/3*(1-1/101)

A=1/3*100/101

A=300/301

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

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

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

A=1/3.5+1/5.7+1/7.9+...+1/99.101

2A= 2/3.5+2/5.7+2/7.9+...+2/99.101

2A= 1/3-1/5+1/5-1/7-1/7+1/7-1/9+...+1/99-1/101

2A=1/3-1/101=98/303

A=(98/303)/2=49/303

A=97

   666

A=1/3.5 + 1/5.7 + 1/7.9 +...+ 1/99.101

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

=1/2.(1/3-1/101)

=49/303

\(A=1/3.5+1/5.7+1/7.9+…+1/99.101\)

A.2=2/3.5+2/5.7+2/7.9+…+2/99.101

A.2=1/3-1/5+1/5-1/7+1/7-1/9+...+1/99-1/101

Vậy

A.2=1/3-1/101=98/303

A=98/303/2=49/303

Đúng không

A = 1/15 + 1/35 + 1/63 + 1/99 + ... + 1/9999

   = 1/3x5 + 1/5x7 + 1/7x9 + 1/9x11 + ... + 1/99x101

A x 2 = 2/3x5 + 2/5x7 + 2/7x9 + 2/9x11 + ... + 2/99x101

         = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + ... + 1/99 - 1/101

         = 1/3 - 1/101 = 98/303

Vậy A = 98/303 : 2 = 49/303