1+1/3+1/6+1/10+.......+2/Xx(X+1)=1và 1989/1991
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ND
1
NM
Nguyễn Minh Quang
Giáo viên
14 tháng 2 2021
ta có
\(1+\frac{1}{3}+\frac{1}{6}+..+\frac{2}{x\left(x+1\right)}=1+2\left(\frac{1}{2}-\frac{1}{3}\right)+2\left(\frac{1}{3}-\frac{1}{4}\right)+..+2\left(\frac{1}{x}-\frac{1}{x+1}\right)=2-\frac{2}{x+1}\)
Nên ta có
\(2-\frac{2}{x+1}=1+\frac{1989}{1991}\Leftrightarrow\frac{2}{x+1}=\frac{2}{1991}\Leftrightarrow x=1990\)
ND
Nguyễn Đức Trí
VIP
21 tháng 8 2023
\(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=1\dfrac{1989}{1991}\)
\(\Rightarrow2\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{3980}{1991}\)
\(\Rightarrow2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{3980}{1991}\)
\(\Rightarrow2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{3980}{1991}\)
\(\Rightarrow2\left(1-\dfrac{1}{x+1}\right)=\dfrac{3980}{1991}\)
\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{3980}{1991}.\dfrac{1}{2}\)
\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{1990}{1991}\)
\(\Rightarrow\dfrac{1}{x+1}=1-\dfrac{1990}{1991}\)
\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{1991}\)
\(\Rightarrow x+1=1991\)
\(\Rightarrow x=1990\)
NT
4
12 tháng 5 2015
Ta có \(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+......+\frac{2}{x\times\left(x+1\right)}=1\frac{1989}{1991}\)
\(=2+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+......+\frac{2}{x\times\left(x+1\right)}=\frac{3980}{1991}\)
\(=2\times\left(1+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+.....+\frac{1}{x\times\left(x+1\right)}\right)=\frac{3980}{1991}\)
\(=2\times\left(1+\frac{1}{2}-\frac{1}{\left(x+1\right)}\right)=\frac{3980}{1991}\)
\(\Rightarrow2+1-\frac{2}{x+1}=\frac{3980}{1991}\)
\(\Rightarrow3-\frac{2}{x+1}=\frac{3980}{1991}\)
\(\Leftrightarrow\frac{2}{x+1}=3-\frac{3980}{1991}=\frac{1993}{1991}\)
\(\Rightarrow\frac{2\times996.5}{x+1}=\frac{1993}{1991}\)
\(\Rightarrow x+1=1991\)
\(\Leftrightarrow x=1991-1=1990\)
21 tháng 5 2018
đây không phải bn nhân lên 2 lần mà là bn chỉ đổi các ps đó thôi
vậy tại sao bn lại viết 1=2 vậy?
XO
27 tháng 9 2020
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
27 tháng 9 2020
\(2\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{x\left(x+1\right)}\right)=\frac{3980}{1991}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x+1\right)}=\frac{1990}{1991}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1990}{1991}\)
\(1-\frac{1}{x+1}=\frac{1990}{1991}\)
\(\frac{1}{x+1}=1-\frac{1990}{1991}\)
\(\frac{1}{x+1}=\frac{1}{1991}\)
\(x+1=1991\)
\(x=1990\)
DA
0
ta có
\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}\) \(=\)\(1+2\)\(\left(\frac{1}{2}-\frac{1}{3}\right)+2\left(\frac{1}{3}-\frac{1}{4}\right)+...+2\left(\frac{1}{x}-\frac{1}{x+1}\right)\)\(=2-\frac{2}{x+1}\)
Nên ta có
\(2-\frac{2}{x+1}=1+\frac{1989}{1991}\Leftrightarrow\frac{2}{x+1}=\frac{2}{1991}\Leftrightarrow x=1990\)