\(A=3\cdot\frac{1}{1\cdot2}-5\cdot\frac{1}{2\cdot3}+7\cdot\frac{1}{3\cdot4}-\cdots+15\cdot\frac{1}{7\cdot8}-17\cdot\frac{1}{8\cdot9}\)

\(=\frac{3}{1\cdot2}-\frac{5}{2\cdot3}+\frac{7}{3\cdot4}-\cdots+\frac{15}{7\cdot8}-\frac{17}{8\cdot9}\)

\(=1+\frac12-\frac12-\frac13+\frac13+\frac14-\cdots+\frac17+\frac18-\frac18-\frac19\)

\(=1-\frac19=\frac89\)

áp dụng công thức này là làm được bạn ạ:

\(\frac{a}{b.c}\) =\(\frac{a}{b}-\frac{a}{c}\)

1/1*2+1/2*3+1/3*4+...+1/9*10

=1-1/2+1/2-1/3+1/3-1/4+...+1/9-1/10

=1-1/10

=9/10

nho k cho minh voi nhe

\(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ ......... + \(\frac{1}{7.8}\)+ \(\frac{1}{8.9}\)+ \(\frac{1}{9.10}\)

\(=\)\(1\)\(-\)\(\frac{1}{10}\)

\(=\)\(\frac{9}{10}\)

9/10 nha

A = ( -4/5 + 4/3 ) + (-5/4 + 14/5) - 7/3

  =   8/15 + 31/20 - 7/3

  =  25/12 - 7/3

  = -1/4

B = 8/3 x 2/5 x 3/8 x 10x 19/92

   = 16/15 x 15/4 x 19/92

   = 4x19/92

   = 19/23

C = - \(\dfrac{5}{7}\) x \(\dfrac{2}{11}\) + \(\dfrac{-5}{7}\) x \(\dfrac{9}{14}\) + \(\dfrac{1}{57}\)

   = - \(\dfrac{10}{77}\) - \(\dfrac{45}{98}\) + \(\dfrac{1}{57}\)

  = - \(\dfrac{635}{1078}\) + \(\dfrac{1}{57}\)

 = - \(\dfrac{36195}{61446}\) + \(\dfrac{1078}{61446}\)

 = - \(\dfrac{35117}{61446}\)

D = \(\dfrac{12}{19}\) x \(\dfrac{7}{15}\) x \(\dfrac{-13}{17}\) x \(\dfrac{19}{12}\)  x \(\dfrac{17}{13}\)

D =  (\(\dfrac{12}{19}\) x \(\dfrac{19}{12}\)) x (- \(\dfrac{13}{17}\) x \(\dfrac{17}{13}\)) x \(\dfrac{7}{15}\)

D =   1  x  (- 1) x \(\dfrac{7}{15}\)

D = -1 x \(\dfrac{7}{15}\)

D = - \(\dfrac{7}{15}\)

Mình không thể giải thích được nhưng kết quả chắc chắn là : \(\frac{8}{9}\)

đặt A=1/1×2+1/2×3+1/3×4+

bn đặt ra đi sau đó rồi tính vd 1/1x2=1/1-1/2

Bài 1:

Đặt \(A=\frac{2}{1x2}+\frac{2}{2x3}+\frac{2}{3x4}+...+\frac{2}{18x19}+\frac{2}{19x20}\)

\(\frac{A}{2}=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{18x19}+\frac{1}{19x20}\)

\(\frac{A}{2}=\frac{2-1}{1x2}+\frac{3-2}{2x3}+\frac{4-3}{3x4}+...+\frac{19-18}{18x19}+\frac{20-19}{19x20}\)

\(\frac{A}{2}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}=1-\frac{1}{20}=\frac{19}{20}\)

\(A=\frac{2x19}{20}=\frac{19}{10}\)

Bài 2:

Đặt \(B=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{8x9}+\frac{1}{9x10}\)

Làm tương tự câu 1 có \(B=1-\frac{1}{10}=\frac{9}{10}\)

\(Bx100=\frac{9}{10}x100=90\)

=> \(\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=1\)

=> \(\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]=\frac{1}{2}\)

=>  \(x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}=5\Rightarrow x=5-\frac{206}{100}=\frac{294}{100}=\frac{147}{50}\)

bài 1 đáp án là:19/10

2:147/50

đáp án bài 1 là: 19/10

đáp án bài 2 là 147/50

a) 3/7 + 4/9 + 4/7 + 5/9

= ( 3/7 + 4/7 ) + ( 4/9 + 5/9 )

= 7/7 + 9/9

= 1  + 1 

= 2

b)1/5 + 4/10 + 9/15 + 16/20 + 25/25 + 36/30 + 49/35 + 64/40 + 81/45

= 1/5 + 2/5 + 3/5 + 4/5 + 5/5 + 6/5 + 7/5 + 8/5 + 9/5

= ( 1/5 + 9/5 ) + ( 2/5 + 8/5 ) + (7/5 + 3/5 ) + ( 4/5 + 6/5  ) + 5/5

= 2 + 2 + 2 + 2 + 1

= 2  x 4 + 1

= 8  +1 

= 9

c)  1/8 + 1/12 + 3/8 + 5/12

= ( 1/8 + 3/8 ) + ( 1/12 + 5/12)

= 4/8 + 6/12

= 1/2 + 1/2

= 2/4 = 1/2

mỏi tay rồi

d; (1 - \(\dfrac{1}{2}\)) x (1 - \(\dfrac{1}{3}\)) x (1 - \(\dfrac{1}{4}\)) x ... x ( 1 - \(\dfrac{1}{100}\))

 = \(\dfrac{1}{2}\) x \(\dfrac{2}{3}\)  x \(\dfrac{3}{4}\) x \(\dfrac{3}{4}\) x ... x \(\dfrac{99}{100}\)

= \(\dfrac{1}{100}\)

e; \(\dfrac{9}{5}\) : \(\dfrac{17}{15}\) + \(\dfrac{8}{5}\) : \(\dfrac{17}{15}\)

=  \(\dfrac{9}{5}\) x \(\dfrac{15}{17}\) + \(\dfrac{8}{5}\) x \(\dfrac{15}{17}\)

=  \(\dfrac{27}{17}\) + \(\dfrac{24}{17}\)

=  \(\dfrac{51}{17}\)

=   3

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{9.10}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}\)

\(=\frac{9}{10}\)

chúc bn học tốt

\(1,27+2,77+4,27+5,77+...+31,27+32,47\)

\(=\left(1,27+32,77\right)+\left(2,77+31,27\right)+....+\left(16,27+17,77\right)\)

\(=34,04+34,04+....+34,04\)( 11 số hạng)

\(=34,04.11=374,44\)

chúc bn học tốt

 b)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}\)

\(=\frac{10}{10}-\frac{1}{10}\)

\(=\frac{9}{10}\)