1/3 + 1/6 + 1/10 + ... + 2/x(x + 1) = 1999/2001
2 × (1/6 + 1/12 + 1/20 + ... + 1/x(x + 1) = 1999/2001
1/2×3 + 1/3×4 + 1/4×5 + ... + 1/x(x + 1) = 1999/2001 : 2
1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/x - 1/x+1 = 1999/2001 × 1/2
1/2 - 1/x+1 = 1999/4002
1/x+1 = 1/2 - 1999/4002
1/x+1 = 2/4002 = 1/2001
=> x + 1 = 2001
=> x = 2001 - 1 = 2000
Vậy x = 2000
\(\frac{1}{3}\)+\(\frac{1}{6}\)+\(\frac{1}{10}\)+........+\(\frac{2}{x\left(x+1\right)}\)=\(\frac{1999}{2001}\)
tìm x tuộc N :
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
Tìm \(x\in N\)biết:
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
a) \(\left(\frac{6^3-10,5^3}{6^2.3^3-15^2.5^2}.\left|x-2\right|\right):10=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{9}\right).\left(1-\frac{1}{10}\right)\)
b) \(\frac{x-2018}{2}+\frac{x-2020}{4}=\frac{x-2040}{8}+\frac{x-2030}{14}\)
Tính
A bằng 1+2-3-4+5+6-7-8+.......-1999-2000+2001+2002-2003
B bằng \(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)\left(\frac{1}{25}-1\right)......\left(\frac{1}{121}-1\right)\)
tìm x biết :
\(\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}\right)+1\)
\(4\left(\frac{1}{2}-x\right)-5\left(x-\frac{3}{10}\right)=\frac{7}{4}\)
\(2\left(\frac{3}{2}-x\right)-\frac{1}{3}=7x-\frac{1}{4}\)
\(-\frac{3}{2}\left(5-\frac{1}{6}\right)+4\left(x-\frac{1}{2}\right)=1\)
Tìm x :
a) \(\frac{x+1}{2000}+\frac{x+2}{1999}+\frac{x+ 3}{1998}+\frac{x+4}{1997}=-4\)
\(b.\frac{x+1}{1999}+\frac{x+2}{2000}+\frac{x+3}{2001}=\frac{x+4}{2002}+\frac{x+5}{2003}+\frac{x+6}{2004}\)
\(\left(\frac{1}{7}x-\frac{2}{7}\right).\left(-\frac{1}{5}x+\frac{3}{5}\right).\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)0
b) \(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{5}x+1=0\)
