học trường gì

\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{99\cdot101}\)

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

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

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

Tích mk nha bn !!!! ^_^

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

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

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

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

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

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

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

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

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

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

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

= 1/3 - 1/101

= 101/303 - 3/303

= 98/303

2 con 2/ 5x7 và 2/7x5 là sao :/

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

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

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

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

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 !!!

Ta có: \(A=\frac{2^2}{3\times5}+\frac{2^2}{5\times7}+...+\frac{2^2}{99\times101}\)

\(\Rightarrow A=2.\)< \(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{99\times101}\)>

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

\(\Rightarrow A=2.< \frac{1}{3}-\frac{1}{101}>\)

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

\(\Rightarrow A=\frac{196}{303}\)

Nhớ k nhá.

M=2/3.5+2/5.7+/.............+2/97/99

=1/3-1/5+1/5-1/7+......+1/97-1/99

=1/3-1/99

=32/99

OK

xin 5 tích vì mất 5 phút

\(A=\frac{3-1}{1.3}+\frac{5-3}{3.5}+...+\frac{101-99}{99.101}\)

\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}=\frac{100}{101}\)

Ta có:

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

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

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

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)

giup mik voi mn nhanh len nhe 

đề sai bạn ơi