tao đét biết 

Tui biết nhưng sợ sai

1 + 2 + 3 + ... + 2001 + 2002

Ta có : 1 + 2 + 3 + ... + 2001 + 2002 ( có 2002 số )

      = (2002 + 1) . 2002 : 2

       = 2005003

Số số hạng là : ( 2002 - 1 ) : 1 + 1 = 2002 ( số )

Tổng là : ( 2002 + 1 ) . 2002 : 2 = 2005003

Vậy,..........

giúp mình với các bạn ơi

3075/100+1/12+3-1/6

=123/4₍₃₎+1/12(1)+3₍₁₂₎+1/6(2)

=369/12+1/12+36/12+2/12

369+1+36+2

12 

=408/12=34

1)Ta có: A= 2004/2005=1- 1/2005          B=2005/2006=1- 1/2006        1/2005>1/2006  =>1- 1/2005 < 1- 1/2006

Vậy A<B.

2)Tương tự như trên,1001/1002<1002/1003

\(\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)

\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{303}{1540}\)

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

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

\(\Rightarrow x+3=308\)

\(\Rightarrow x=305\)

a) \(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x.\left(x+3\right)}=\frac{101}{1540}\)

\(\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{101}{1540}\)

\(\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

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

\(\frac{1}{5}-\frac{1}{x+3}=\frac{101}{1540}:\frac{1}{3}\)

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

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

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

\(\Rightarrow x+3=308\)

\(\Leftrightarrow x=308-3\)

\(\Leftrightarrow x=305\)

Vậy \(x=305\)