mình nghĩ bạn điền sai đề

ukm mk đánh sai thiệt phải là 1/5x6 chứ ko phải 1/4x5

Sai đề à bạn    

Bài này khi sáng mình mới học 100% là đúng luôn.

1/1x2 + 1/2x3 + 1/3x4 + 1/4x5 + ........... + 1/99x100.

=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+..........1/98-1/99+1/99-1/100.

=1/1-1/100=99/100.

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)

1/1x2 + 1/2x3 + 1/3x4 + 1/4x5 + ........... + 1/99x100.

=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+..........1/98-1/99+1/99-1/100.

=1/1-1/100

=99/100

Có \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)

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

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

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

=\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)

= \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

=> \(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right):\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=1\)

B x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3

= 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)

= 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.

= 99x100x101

B = 99x100x101 : 3

= 333300

nhanh k minh

B= 1x2+3x4+5x6+...+99x100

=> Bx3= 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ...+ 99x100x3

=> Bx3= 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3)+...+99x100x(101-98)

=> Bx3= 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 +4x5x6 - 3x4x5 +...+ 99x100x101 - 98x99x100

=> Bx3= 99x100x101

=> B= 99x100x101:3

=> B= 333300

ta có:\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

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

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)=\frac{1}{51}+...+\frac{1}{100}\)

\(\frac{2012}{51}+\frac{2012}{52}+...+\frac{2012}{100}=2012\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)

bài toán được viết lại như sau:

\(\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right).x=2012\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)

\(\Rightarrow x=2012\left(\frac{1}{51}+...+\frac{1}{100}\right):\left(\frac{1}{51}+...+\frac{1}{100}\right)\)

\(\Rightarrow x=2012\)

vậy x=2012

B = 1/1x2 + 1/3x4 + ... + 1/99x100

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

B = (1 + 1/2 + 1/3 + 1/4 + ... + 1/99 + 1/100) - (2.1/2 + 2.1/4 + 2.1/6 + ... + 2.1/100)

B = (1 + 1/2 + 1/3 + 1/4 + ... + 1/99 + 1/100) - (1 + 1/2 + 1/3 + ... + 1/50)

B = 1/51 + 1/52 + 1/53 + ... + 1/100

=> tỉ số a/b = 1

1/1x2+1/2x3+1/3x4+..+1/9x10

=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5+...-1/10

=1-1/10

=9/10