B=2/3x5 + 2/5x7 + 2/7x9 + ...+2/99x101

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

B= 1/3 - 1/101

B=98/303

( k mk nhé ! Cách làm câu a và b của mk đều đúng 100% đấy ! Dạng này mk học từ lâu rồi ! )

a, A = 1/2x3+ 1/ 3x4 + 1/4x5 + 1/5x6 + ... + 1/99x100

    A= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 -1/5 + 1/5 - 1/6 + ... + 1/99 -1/100

    A= 1/2 -1/100

    A= 49 / 100

nhờ bạn giải dùm mình câu c,d

Ta có:

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

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

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

Vậy \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}< 1\)

Đặt biểu thức là A 

Ta có A = (3-1)1x3 + (5-3)/3x5+..........+(101-99)/101x99

            =3/1x3 - 1/1*3 + 5/3x5 - 3/3x5 + ...........+ 101/99x101 - 99/101x99

            = 1- 1/3 +1/3 -1/5 +............+ 1/99 - 1/101

              = 1 -1/101 < 1 (Điều phải chứng minh)

a)\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)

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

= \(1-\dfrac{1}{101}\)

=\(\dfrac{100}{101}\) 

\(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{5.7}+...+\dfrac{5}{99.101}\)

=\(\dfrac{5}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99+101}\right)\)

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

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

= \(\dfrac{5}{2}-\dfrac{100}{101}\)

= \(\dfrac{305}{202}\)

Bài 16: 

A = \(\dfrac{2n+1}{3n+1}\); đkxđ n \(\ne\) - \(\dfrac{1}{3}\)

Gọi ước chung lớn nhất của 2n + 1 là d

Ta có: 2n + 1 ⋮ d; 3n + 1 ⋮ d

2n + 1 ⋮ d ⇒ 3.(2n + 1) ⋮ d ⇒ 6n + 3 ⋮ d

3n + 1 ⋮ d ⇒ 2.( 3n+ 1) ⋮ d ⇒ 6n + 2 ⋮ d

 ⇒ 6n + 3 - (6n + 2) ⋮ d ⇒ 6n + 3 - 6n - 2⋮ d

                                     ⇒ 1 ⋮ d ⇒ d = 1

Ước chung lớn nhất của 2n + 1 và 3n + 1 là 1 

Hay 2n + 1 và 3n + 1 là hai số nguyên tố cùng nhau (đpcm)

\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+...+\dfrac{2}{99\times101}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ =1-\dfrac{1}{101}\\ =\dfrac{100}{101}\)

= 100/101

Ta có : \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\)

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

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

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

Đặt : \(A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)

\(A-\frac{2}{1\cdot3}=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)

\(2A-\frac{2}{1\cdot3}=\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-...+\frac{2}{99}-\frac{2}{101}\)

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

\(2A-\frac{2}{3}=\frac{196}{303}\)

\(A-\frac{2}{3}=\frac{98}{303}\)

\(A=\frac{98}{303}+\frac{2}{3}=\frac{100}{101}\)

\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{99\times101}\)

\(=\frac{3-1}{1\times3}+\frac{5-3}{3\times5}+\frac{7-5}{5\times7}+...+\frac{101-99}{99\times101}\)

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

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

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

A = 2/1x3 + 2/3x5 + 2/5x7 + ... + 2/99x101

A = 2/1 - 2/101 = 200/101

Kết quả là 200/101 bạn nhé

2/2 + 1x3 / 3x5 + 2/2 + ······ + 5x7 / 97x99 + 2 / 99x101 
= 1-1 / 3 + ​​1 / 3-1 / 5 + 1 / 5-1 / 7 + ... ... + 1 / 97-1 / 99 + 1 / 99-1 / 101 
= 1-1 / 101 
= 100/101

chưa chắc là 200/101

\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)

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

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

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

2/1.3 + 2/3.5 + 2/5.7 + ... + 2/99.101

= 2 .( 1/1.3 + 1/3.5 + 1/5.7 + ... + 1/99.101 ) 

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

= 2 . ( 1 - 1/101 ) 

= 2 . ( 101/101 - 1/101 ) 

= 2 . 100/101

= 200/101

Chúc bn hok tốt !!!

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\)

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

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

\(=\frac{1}{2}.\frac{100}{101}\)

\(=\frac{50}{101}\)

\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{99\cdot101}\)

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

\(=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{99\cdot101}\)

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

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