Ta có :\(B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}.....\frac{98^2}{98.99}=\frac{\left(1.2.3.4...98\right).\left(1.2.3.4...98\right)}{\left(1.2.3.4...98\right).\left(2.3.4.5...99\right)}=\frac{1}{99}\)

Lại có A = \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}\)

Lại có \(A:B=\frac{98}{99}:\frac{1}{99}=98\)

=> A = 98B

các bạn có  về sweet home

a, \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{44}{45}\)

=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{44}{45}\)

=> \(1-\frac{1}{x+1}=\frac{44}{45}\)

=> \(\frac{x}{x+1}=\frac{44}{45}\)

=> x = 44

b, Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)

.................

\(\frac{1}{45^2}< \frac{1}{44.45}=\frac{1}{44}-\frac{1}{45}\)

=> \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{44}-\frac{1}{45}=1-\frac{1}{45}< 1\)

Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{45^2}< 1\)

a) 1/1.2+1/2.3+1/3.4+...+1/x(x+1)=1-1/2+1/2-1/3+1/3-1/4+....+1/x-1/(x+1)=1-1/(x+1)=x/(x+1)=44/45

=> x=44

b/ 1/2² < 1/1.2; 1/3² < 1/2.3; ....; 1/45² < 1/44.45

=> A < 1/1.2+1/2.3+...+1/44.45=1-1/45=44/45 < 1

=> A < 1

giải hộ mình với cảm ơn

Bài 2: 

a: 

=>-2x=-6

hay x=3

bài 1

a) 196:4-12.(-5)=49-(-60)=109

b)-37.(1-382)-(382.37)= -37.(-381)-14134=14097-14134=-37

d) Ta có: \(x+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{-37}{45}\)

\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{-37}{45}\)

\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{-37}{45}\)

\(\Leftrightarrow x=\dfrac{-37}{45}+\dfrac{1}{45}-\dfrac{1}{5}=\dfrac{-36}{45}-\dfrac{1}{5}=\dfrac{-4}{5}-\dfrac{1}{5}=-1\)

Vậy: x=-1

1)

2/3.5+2/5.7+...+2/11.13+2/13.15+2/1.2+2/2.3+...+2/9.10

=(2/3.5+...2/13.15)+(2/1.2+...+2/9.10)

= (2/3-2/15)+ [2(1-1/10)]

=8/15+9/5

=7/3

2)

11/12+11/12.24+...+11/88.99

=11-1/9

=10/8/9

a) \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{99\cdot100}\)

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

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

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

b) \(\frac{2}{3\cdot5}+\frac{3}{5\cdot7}+...+\frac{2}{49\cdot51}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)

\(=\frac{1}{3}-\frac{1}{51}\)

\(=\frac{16}{51}\)

a) 1/1.2+1/2.3+1/3.4+...+1/99.100

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

= 1/1 - 1/100

= 99/100

b) 2/3.5+2/5.7+...+2/49.51

= 2 . ( 1/3.5 + 1/5.7 + ... + 1/49.51 )

= 2 . ( 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/49 - 1/50 )

= 2 . ( 1/3 - 1/50 )

= 2 . 47/150

= 47/75