Bài 2:

\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2004}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}\)

\(=\frac{1}{2004}\)

Bài 2

=1/2 x 2/3 ... x 2003/2004

=1/2004

2.

1/2004

sudy well

study well

Mình không viết lại đề bài nha

a) \(\Rightarrow\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

\(\Rightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)

\(\Rightarrow\frac{1}{x+3}=\frac{1}{308}\Rightarrow x=305\)

Tìm x,y thuộc Z biết:

a, \(2^{x+y}=2^x+2^y\)

b, \(x+y=x.y=x:y\left(y\ne0\right)\)

Làm nhanh giùm mình nhé!!!!!

ngại làm quá

Mình giải luôn ra nha

Tìm x:
1/3+1/6+1/10+.........+2/x.(x+1)=499/1000
1/2.(1/3+1/6+1/10+.......+2/x.(x+1)=499/1000.1/2
1/6+1/12+1/20+.......+1/x.(x+1)=499/2000
1/(2.3)+1/(3.4)+1/(4.5)+.........+1/x.(x+1)=499/2000
1/2-1/3+1/3-1/4+1/4-1/5+........+1/x-1/(x+1)=499/2000
1/2-1/(x+1)=499/2000
1/(x+1)=1/2-499/2000
1/(x+1)=501/2000
\Rightarrow1.2000=(x+1).501
\Rightarrow2000=x.501+501
\Rightarrow1499=x.501
\Rightarrowx=1499:501
Vì x thuộc Z nên 1499:501 là 1 số nguyên.Mà 1499:501 được 1 số thập phân nên x thuộc rỗng.

1.

\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)

\(MC:12\)

Quy đồng :

\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)

\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)

\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)

\(\Leftrightarrow6x+9-3x=-4-9+16\)

\(\Leftrightarrow-7x=3\)

\(\Leftrightarrow x=\frac{-3}{7}\)

2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)

\(MC:20\)

Quy đồng :

\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)

\(\Leftrightarrow30x+15-20=15x-2\)

\(\Leftrightarrow15x=3\)

\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)

Đặt A=1/3+1/6+1/10+...+2/x*(x+1)

        1/2A=1/3*2+1/6*2+1/10*2+...+2/2*x*(x+1)

         1/2A=1/6+1/12+1/20+...+1/x*(x+1)

          1/2A=1/2*3+1/3*4+1/4*5+...+1/x*(x+1)

           1/2A=1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/(x+1)

           1/2A=1/2-1/x+1

           A=(1/2-1/x+1):1/2

          A=1-2/x+1

Ta có A=1999/2001

Hay 1-2/x+1=1999/2001

           2/x+1=1-1999/2001

          2/x+1=2/2001

=>x+1=2001

=>x=2000

Cho A = 1/3+1/6+1/10+...+2/x(x+1)

    1/2A= 1/3.2+1/6.2+1/10.2+...+2/x(x+1)2

    1/2A= 1/6+1/12+1/20+...+1/x(x+1)

    1/2A= 1/2.3+1/3.4+1/4.5+...+1/x(x+1)

    1/2A= 1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1

    1/2A= 1/2-1/x+1

    A      = (1/2-1/x+1)/1/2

    A      = 1-2/x+1

Mà A=1999/2001

=> 1-2/x+1= 1999/2001

         2/x+1= 1-1999/2001

         2/x+1= 2/2001

     =>x+1=2001

     =>x     = 2000

Đặt N=1/10+1/15+1/21+...+2/x*(x+1)

1/2N=1/20+1/30+1/42+...+1/x*(x+1)

1/2N=1/4*5+1/5*6+1/6*7+...+1/x*(x+1)

1/2N=1/4-1/5+1/5-1/6+1/6-1/7+...+1/x-1/x+1

1/2N=1/4-1/x+1

N=(1/4-1/x+1):1/2

N=1/2-2/x+1

Thiếu đề

x=2009 dễ mà

mk làm câu c cho nó dễ

c)1/1.2+1/2.3+...+1/x.(x+1)=2009/2010

=1-1/2+1/2-1/3+...+1/x-1/x+1=2009/2010

=1-1/x+1=2009/2010

=1/x+1=1-2009/2010

=1/x+1=1/2010

=) x+1=2010

x         =2010-1

x         =2009

Đề cho dài :v. Lần sau đăng từ từ nhé bạn, hôm qua đến giờ mình giải không hết đó =(((

a) \(\frac{1}{2}.x-\frac{3}{4}.x-\frac{7}{3}=-\frac{5}{6}=\frac{-5}{6}\)

\(\frac{1}{2}.x-\frac{3}{4}.x=\frac{-5}{6}+\frac{7}{3}=\frac{3}{2}\)

\(\Leftrightarrow x\left(\frac{1}{2}-\frac{3}{4}\right)=\frac{3}{2}\Leftrightarrow x.\frac{-1}{4}=\frac{3}{2}\)

\(x=\frac{3}{2}:\frac{-1}{4}=-6\)

b) \(\frac{4}{5}.x-x-\frac{3}{2}.x+\frac{4}{3}=\frac{1}{2}-\frac{6}{5}=-\frac{7}{10}\)

\(\Leftrightarrow x\left(\frac{4}{5}-\frac{3}{2}.\frac{4}{3}\right)=x\left(\frac{4}{5}-2\right)=-\frac{7}{10}\)

\(\Leftrightarrow x.\frac{-6}{5}=-\frac{7}{10}\)

\(x=-\frac{7}{10}:\frac{-6}{5}=\frac{7}{12}\)

c) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{2009}{2010}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{2010}\)

\(=1-\frac{1}{x+1}=\frac{2009}{2010}\)

\(\frac{1}{x+1}=1-\frac{2009}{2010}=\frac{1}{2010}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2010-1}=\frac{1}{2009}\). Vậy x= 2009

d) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2013}{2015}=\frac{4023}{2015}\)

\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{4023}{2015}\)

\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{4023}{2015}\)

\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{4023}{2015}\)

\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{4023}{2015}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{4023}{2015}:2=\frac{4023}{4030}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{4023}{4030}=\frac{-1004}{2015}=\frac{1004}{-2015}\)

\(x+1=\hept{\begin{cases}2015\\-2015\end{cases}}\Rightarrow x=\hept{\begin{cases}2014\\-2016\end{cases}}\)

e) Bạn tự làm, nhiều quá mình làm không hết

\(a,\left(\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|\right):10=\left(1-\frac{1}{2}\right)....\left(1-\frac{1}{10}\right)\)

\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\Leftrightarrow\frac{6^3-10.5^3}{6^2.3^3-15^2.5^2}.|x-2|=1\)

\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.|x-2|=1\Leftrightarrow|x-2|.\frac{2}{3}=1\Leftrightarrow|x-2|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)

\(\left(\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|\right):10=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{9}\right).\left(1-\frac{1}{10}\right)\)

\(=\frac{1.2.3.4...9}{1.2.....10}=\frac{1}{10}\)

\(\Leftrightarrow\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|=1\)

\(\Leftrightarrow\frac{6^2.6-2.5^4}{6^2.3^2-3^2.5^4}.\left|x-2\right|=1\)

\(\Leftrightarrow\left|x-2\right|.\frac{2}{3}=1\Leftrightarrow\left|x-2\right|=\frac{3}{2}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)

Mình làm tiếp câu b nha !

b,                                               Bài giải

\(\frac{x-2018}{2}+\frac{x-2020}{4}=\frac{x-2040}{8}+\frac{x-2030}{14}\)

\(\left(\frac{x-2018}{2}+1\right)+\left(\frac{x-2020}{4}+1\right)=\left(\frac{x-2040}{8}+1\right)+\left(\frac{x-2030}{14}+1\right)\)

\(\frac{x-2016}{2}+\frac{x-2016}{4}=\frac{x-2032}{8}+\frac{x-2016}{14}\)

\(\left(x-2016\right)\left(\frac{1}{2}+\frac{1}{4}\right)=\frac{x-2016}{8}-2+\frac{x-2016}{14}\)

\(\left(x-2016\right)\cdot\frac{3}{4}=\left(x-2016\right)\left(\frac{1}{8}+\frac{1}{14}\right)-2\)

\(\left(x-2016\right)\cdot\frac{3}{4}=\left(x-2016\right)\cdot\frac{11}{56}-2\)

\(\left(x-2016\right)\cdot\frac{3}{4}-\left(x-2016\right)\cdot\frac{11}{56}=-2\)

\(\left(x-2016\right)\left(\frac{3}{4}-\frac{11}{56}\right)=-2\)

\(\left(x-2016\right)\cdot\frac{31}{56}=-2\)

\(x-2016=-2\text{ : }\frac{31}{56}\)

\(x-2016=-\frac{112}{31}\)

\(x=-\frac{112}{31}+2016\)

\(x=\frac{62384}{31}\)