Tu de bai ta co
1/6+1/12+1/20+...+1/(x*(X+1))=1999/4002
Suy ra 1/(2*3)+1/(3*4)+1/(4*5)+...+1/(x*(x+1))=1999/4002
Suy ra 1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1=1999/4002
Suy ra 1/2-1/(x+1)=1999/4002
Suy ra 1/(x+1)=1/2001
Suy ra x+1=2001
Suy ra x=2000
Tìm x biết
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
TÌm x, biết: \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right)}=\frac{1999}{2001}\)
tìm số nguyên x biết rằng
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{1999}{2001}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{1999}{2001}\)
Tìm x
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{2001}\)
Tìm x.
Tìm số tự nhiên x biết rằng:
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{2001}\)
Tìm số tự nhiên x biết rằng
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{2001}\)
Giải đầy đủ nha, ko sai đề đâu
Tìm x : \(1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}=1\frac{1999}{2001}\)
Tìm x, biết :
a, \(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{98\cdot99\cdot100}\right)x=-3\);
b, \(\left(\frac{\frac{2000}{1}+\frac{1999}{2}+...+\frac{1}{2000}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}}\right)x=\frac{-1}{5}\).
c,\(\left(\frac{\frac{2000}{1}+\frac{1999}{2}+...+\frac{1}{2000}+2000}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}}\right):x=\frac{-2001}{2002}\).