Tính : 1 + \(\frac{1}{3}\) +\(\frac{1}{6}\)+\(\frac{1}{10}\)+...+\(\frac{2}{x\left(x+1\right)}\)= \(1\frac{1989}{1991}\)
Giúp mình nhóe
1+\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{2}{x\left(x+1\right)}=1\frac{1989}{1991}\)
Ta có : \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{1989}{1991}\)
=> \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{1989}{1991}\)
=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{1989}{1991}\)
=> \(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1989}{1991}\)
=> \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1989}{3982}\)
=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1989}{3982}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{1989}{3982}\)
=> \(\frac{1}{x+1}=\frac{1}{1991}\)
=> x + 1 = 1991
=> x = 1990
Vậy x = 1990
\(2\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{x\left(x+1\right)}\right)=\frac{3980}{1991}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}=\frac{1990}{1991}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1990}{1991}\)
\(1-\frac{1}{x+1}=\frac{1990}{1991}\)
\(\frac{1}{x+1}=1-\frac{1990}{1991}\)
\(\frac{1}{x+1}=\frac{1}{1991}\)
\(x+1=1991\)
\(x=1990\)
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+......+\frac{2}{x\left(x+1\right)}=1\frac{1989}{1991}\)
\(\Leftrightarrow\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.......+\frac{2}{x\left(x+1\right)}=\frac{1989}{1991}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+.......+\frac{2}{x\left(x+1\right)}=\frac{1989}{1991}\)
\(\Leftrightarrow2.\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+......+\frac{1}{x\left(x+1\right)}\right]=\frac{1989}{1991}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{x\left(x+1\right)}=\frac{1989}{3982}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.......+\frac{1}{x\left(x+1\right)}=\frac{1989}{3982}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{1989}{3982}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1989}{3982}\)\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{1991}\)
\(\Leftrightarrow x+1=1991\)\(\Leftrightarrow x=1990\)
Vậy \(x=1990\)
\(2+\frac{2}{3}+\frac{2}{6}+....+\frac{2}{x\left(x+1\right)}=1\frac{1989}{1991}\)
Hình như đề sai rồi bạn ạ
Bạn xem lị đề nha
Sai đề rồi bạn ơi, 2 + ... không thể nào = 1 1989/1991 được bạn ạ !!!
đề thầy giáo ra mà
thầy dạy HSG lớp 9 đó
Tìm x
\(2+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+......+\frac{2}{x\left(x+1\right)}=1\frac{1989}{1991}\)
\(2+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=1\frac{1989}{1991}\)
\(2\left(1+\frac{1}{3}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=1\frac{1989}{1991}\)
\(2\left(1+\frac{1}{3}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=1\frac{1989}{1991}\)
\(2\left(1+\frac{1}{3}+\frac{1}{2}-\frac{1}{x+1}\right)=1\frac{1989}{1991}\)
\(\frac{8}{3}+2-\frac{2}{x+1}=1\frac{1989}{1991}\)
\(\frac{2}{x+1}=\frac{13}{10}\)( số thập phân dài quá nên mk lấy số tròn thôi nha )
\(x+1=2:\frac{13}{10}\)
\(x+1=\frac{20}{13}\)
\(\Leftrightarrow x=\frac{7}{13}\)
1+\(\frac{2}{6}+\frac{2}{12}+......+\frac{2}{x.\left(x+1\right)}=1\frac{1989}{1991}\)
Tìm x, biết:
\(2+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+.....+\frac{2}{x\left(x+1\right)}=1\frac{1989}{1991}\)
\(2\left(1+\frac{1}{3}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{x\left(x+1\right)}\right)=\frac{3980}{1991}\)
\(1+\frac{1}{3}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+......+\frac{x+1-x}{x\left(x+1\right)}=\frac{1990}{1991}\)
\(1+\frac{1}{3}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{x}-\frac{1}{x-1}=\frac{1990}{1991}\)
\(1+\frac{1}{3}+\frac{1}{2}-\frac{1}{x-1}=\frac{1990}{1991}\)
\(\frac{1}{x-1}=\frac{11}{6}-\frac{1990}{1991}=\frac{9961}{11946}\)
\(x-1=\frac{11946}{9961}\Rightarrow x=\frac{21907}{9961}\)
\(y=2+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\cdot\left(x+1\right)}=1\frac{1989}{1991}\)
giúp mình với
1+ \(\frac{1}{3}\) + \(\frac{1}{6}\)+ \(\frac{1}{10}\) +...+ \(\frac{2}{x\left(x+1\right)}\) = \(1\frac{1989}{1991}\)
tìm x:
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{x.\left(x+1\right):2}=\frac{1991}{1993}\)
Tính tổng :
\(\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{3^2}\right)+\left(1-\frac{1}{4^2}\right)+....+\left(1-\frac{1}{50^2}\right)\)
Câu 1:
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}=\frac{1991}{1993}.\)
\(\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{x.\left(x+1\right):2}=\frac{1991}{1993}\)
\(\frac{1}{2.3}.2+\frac{1}{3.4}.2+\frac{1}{4.5}.2+...+\frac{1}{x.\left(x+1\right)}.2=\frac{1991}{1993}\)
\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{1991}{1993}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1991}{3986}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1991}{3986}\)
...
e tự tính nốt nha
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{1991}{1993}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{1991}{1993}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1991}{1993}\div2\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1991}{3986}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1991}{3986}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1991}{3986}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1991}{3986}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{1993}\)
\(\Leftrightarrow x+1=1993\)
\(\Leftrightarrow x=1993-1\)
\(\Leftrightarrow x=1992\)
Vậy x = 1992
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+............+\frac{1}{x\left(x+1\right):2}=1\frac{1991}{1993}\)
Tính tổng
=> 1/2+1/6+1/12+1/20+....+1/x.(x+1) = 1992/1993
=> 1/2+1/2.3+1/3.4+1/4.5+.....+1/x.(x+1) = 1992/1993
=> 1/2+1/2-1/3+1/3-1/4+1/4-1/5+.....+1/x-1/x+1 = 1992/1993
=> 1 - 1/x+1 = 1992/1993
=> x/x+1 = 1992/1993
=> x = 1992
Vậy x = 1992
Tk mk nha
\(\Rightarrow\frac{2}{2}+\frac{2}{2.3}+\frac{2}{2.6}+...+\frac{2}{x\left(x+1\right)}=\frac{3984}{1993}\)
\(\Rightarrow2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{3984}{1993}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{3984}{1993}:2\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{1992}{1993}\)
\(\Rightarrow\frac{x}{x+1}=\frac{1992}{1993}\)
=>x=1992
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