làm ơn hãy giúp mình

Bài 1:

$M=3.4.5+4.5.6+...+13.14.15$

$4M=3.4.5(6-2)+4.5.6(7-3)+....+13.14.15(16-12)$

$=-2.3.4.5+3.4.5.6-3.4.5.6+4.5.6.7+....-12.13.14.15+13.14.15.16$

$=-2.3.4.5+13.14.15.16=43560$

$M=43560:4=10890$

Bài 2:

a.

$3M=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{97.100}$

$=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{100-97}{97.100}$

$=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}$

$=1-\frac{1}{100}=\frac{99}{100}$

$M=\frac{99}{100}:3=\frac{33}{100}$

(1-1/2)(1-1/3)(1-1/4)….(1-1/2002).x=1-1/1x2-1/2x3-1/3x4-...1/2002x2003 ae ghi lời giải jup mình nhé. Tìm x

Gọi \(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)....\left(1-\frac{1}{2002}\right).x\)

\(\Rightarrow A=\frac{1}{2}.\frac{2}{3}....\frac{2001}{2002}.x=\frac{x}{2002}\)

Gọi \(B=1-\frac{1}{1.2}-\frac{1}{2.3}-...-\frac{1}{2002.2003}\)

=>\(B=1-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2002.2003}\right)\)

\(\Rightarrow B=1-\left(1-\frac{1}{2003}\right)=1-\frac{2002}{2003}=\frac{1}{2003}\)

\(\Rightarrow\frac{x}{2002}=\frac{1}{2003}\Rightarrow x=\frac{2002}{2003}\)

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 \(\dfrac{99}{100}\)

= \(\dfrac{1}{100}\)

1) 1/1.2 + 1/2.3 + ... + 1/6.7

= 1 - 1/2 + 1/2 - 1/3 + ... + 1/6 - 1/7

= 1 - 1/7

= 6/7

2) 1/2 + 1/6 + 1/12 + .. + 1/72

= 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9

= 1 - 1/9

= 8/9

3) \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2019}\right)\)

= \(\frac{1}{2}.\frac{2}{3}...\frac{2019}{2020}\)

= \(\frac{1.2....2019}{2.3...2020}\)

= \(\frac{1}{2020}\)

4) A = \(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{512}\)

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

=> 2A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\)

Lấy 2A - A = \(\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^9}\right)\)

             A  = \(\frac{1}{2}-\frac{1}{2^9}\)

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 \(\dfrac{9}{14}\) + \(\dfrac{1}{57}\)

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

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

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

 = - \(\dfrac{35117}{61446}\)

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

\(< =>\frac{128}{256}+\frac{64}{256}+\frac{32}{256}+\frac{16}{256}+\frac{8}{256}+\frac{4}{256}+\frac{2}{256}+\frac{1}{256}\)

\(< =>\frac{128+64+32+16+8+4+2+1}{256}\)

\(< =>\frac{255}{256}\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

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

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

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

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

\(< =>\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)

\(< =>\frac{1\cdot2\cdot3\cdot...\cdot99}{2\cdot3\cdot4\cdot...\cdot100}\)

\(< =>\frac{1}{100}\)

mk chuc ban hoc tot nhe :))