Ta có: \(\left(\frac{x+4}{2000}\right)+\left(\frac{x+3}{2001}\right)=\left(\frac{x+2}{2002}\right)+\left(\frac{x+1}{2003}\right)\)

\(\Rightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)

\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)

\(\Rightarrow\left(x+2004\right).\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}+\frac{1}{2003}\right)=0\)

Vì \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}+\frac{1}{2003}\ne0\)

=> x + 2004 =0

=> x             = -2004

hê hê!viết lại đề:

x₁+x₂+x₃+...+x₂₀₀₂=0 và x₁+x₂+x₃=x₄+x₅+x₆=....=x₁₉₉₉+x₂₀₀₀+x₂₀₀₁=1

\(\Rightarrow x_1+x_2+x_3+...+x_{1999}+x_{2000}+x_{2001}=667\)

\(\Rightarrow0-x_{2002}=667\)

\(\Rightarrow x_{2002}=-667\)

Ta có; \(\frac13+\frac16+\frac{1}{10}+\cdots+\frac{1}{\frac{x\left(x+1\right)}{2}}=\frac{2001}{2002}\)

=>\(\frac26+\frac{2}{12}+\frac{2}{20}+\cdots+\frac{2}{x\left(x+1\right)}=\frac{2001}{2002}\)

=>\(2\left(\frac16+\frac{1}{12}+\frac{1}{20}+\cdots+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2002}\)

=>\(2\left(\frac12-\frac13+\frac13-\frac14+\cdots+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2001}{2002}\)

=>\(2\left(\frac12-\frac{1}{x+1}\right)=\frac{2001}{2002}\)

=>\(1-\frac{2}{x+1}=\frac{2001}{2002}\)

=>\(\frac{2}{x+1}=1-\frac{2001}{2002}=\frac{1}{2002}\)

=>x+1=4004

=>x=4003(nhận)

Ta có A = 1/2×5 -1/5×8 -1/8×11 -1/11×14 -1/14×17 -1/17*20

=>A3= 3/2×5 -3/5×8 -3/8×11 -3/11×14 -3/14×17 -3/17×20

=>A3= 1/2 -1/5 -1/5 +1/8 -1/8 +1/11 -1/11+1/14 -1/14 +1/17 -1/17 +1/20

=>A3= 1/2 -1/5-1/5+1/20

=>A3= 10/20 -4/20 -4/20 +1/20= 3/20

=>A=3/20:3

=> A =1/20 

Có j ko hiu hỏi mk nha

giúp mk bài dưới dc k