Ta xét riêng tử số:

\(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+......+\frac{1}{97}+\frac{1}{99}\)

\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+......+\left(\frac{1}{49}+\frac{1}{51}\right)\)

\(=\frac{100}{1\times99}+\frac{100}{3\times97}+\frac{100}{5\times95}+......+\frac{100}{49\times51}\)

\(=100\times\left(\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+......+\frac{1}{49\times51}\right)\)

Bây giờ xét đến mẫu số:

\(\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+......+\frac{1}{97\times3}+\frac{1}{99\times1}\)

\(=\frac{2}{1\times99}+\frac{2}{3\times97}+\frac{2}{5\times95}+......+\frac{2}{49\times51}\)

\(=2\times\left(\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+......+\frac{1}{49\times51}\right)\)

Vậy giá trị của biểu thức là: \(\frac{100}{2}=50\)

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{98\times99}+\frac{1}{99\times100}\)

\(=\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+\frac{4-3}{3\times4}+....+\frac{99-98}{98\times99}+\frac{100-99}{99\times100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

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

\(=\frac{99}{100}\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}+\frac{1}{99\cdot100}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

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

\(=\frac{99}{100}\)

Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)

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

\(A=\frac{99}{100}\)

Câu A

Ta có (1/2)A  = 1/2² + 1/2³ + ... + 1/2¹⁰⁰ + 1/2¹⁰¹

=> (1/2)A - A = - (1/2)A = (1/2² + 1/2³ + ... + 1/2¹⁰⁰ + 1/2¹⁰¹) - (1/2 + 1/2² + ... + 1/2¹⁰⁰ )

                                   = 1/2¹⁰¹ - 1/2

=> A = 1 - 1/2¹⁰⁰

Câu B

Ta có 1/(1x2) = 1/1 - 1/2

         1/(2.3) = 1/2 - 1/3

  .................................

        1/(99.100) = 1/99 - 1/100

=> B = 1/1 - 1/2 + 1/2 - 1/3 +.... +1/99 - 1/100

        = 1 - 1/100

        =99/100

\(=\frac{1.2}{99.100}\)

\(=\frac{2}{9900}=\frac{1}{4950}\)

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

=>A=\(\frac{7}{2}\)(1-\(\frac{1}{101}\))

=>A=\(\frac{350}{101}\)

7/2 ( \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{99}-\frac{1}{101}\))

7/2 ( 1 - 1/101 ) 

7/2 x 100/101

=350/101 

= 7 phần 1 + 7 phần 3 - 7 phần 3 + 7 phần 5 - 7 phần 5 + 7 phần 7 + ...... + 7 phần 99 - 7 pjaafn 101

= 7 - 7 phần 101 

 = tự tính kết quả nhé , mà bạn viết ra giấy rồi viết vào vở cho dễ nhìn nhé

1/1x2 + 1/2x3 +1/3x4 + ......+1/98x99+1/99x100

=1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +......+ 1/98 - 1/99 + 1/99 + 1/100

=(1-1/100)+(1/2 - 1/2 ) + ( 1/3 - 1/3 ) + ...... + (1/98 - 1/98 ) + ( 1/99 - 1/99 )

= 100/100 - 1/100 + 0 + 0 +.....+ 0 + 0

=99/100

vậy GTBT = 99/100

bn vào câu hỏi tương tự là có

= 1/1 + 1/100

=1011/100

Tích nha

A = 5(1/1.2 + 1/2.3 +......+ 1/99.100)

A = 5( 1 - 1/2 + 1/2 - 1/3 +........+ 1/99 - 1/100)

A = 5( 1 - 1/100)

A = 5 . 99/100

A = 99/20

** k mk nha!

\(\frac{5}{1\times2}+\frac{5}{2\times3}+...+\frac{5}{99\times100}=5\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{99\times100}\right)=5\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=5\left(1-\frac{1}{100}\right)=5\times\frac{99}{100}=\frac{99}{20}=4\frac{19}{20}\)

A=5/1-5/2+5/2-5/3+...+5/99-5/100

A=5-(5/2-5/2+5/3-5/3+...5/99-5/99)-5/100

A=5-0+0+0+...+0-5/100

A=5-5/100

A=49/10=4,9