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
\(2-\frac{2}{x+1}=1+\frac{1989}{1991}\Leftrightarrow\frac{2}{x+1}=\frac{2}{1991}\Leftrightarrow x=1990\)
1+1\3+1/6+1/10+.............+2/x*(x+1)=1 1989/1991
1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\)
+......+ \(\dfrac{2}{x(x+1)}\) =1\(\dfrac{1989}{1991}\)
HeLp me
1 + \(\dfrac{1}{3}\) +\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\) +......+
\(\dfrac{2}{x(x+1)}\) =1\(\dfrac{1989}{1991}\)
\(\dfrac{help}{me}\)
Tìm x biết :1+1/3+1/6+1/10+...+2/x*(x+1)=1/1989/1991 (hỗn số)
2+2/3+2/6+...+2/x*(x+1)=1*1989/1991
2+ 2/3+2/6+2/12+...+2/x(x+1)= 1 1989/ 1991
1+\(\dfrac{1}{3}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+.....+
\(\dfrac{1}{(2x-1)(2x+1)}\)
=1\(\dfrac{1989}{1991}\)
Help meeee
tìm x biết :
2+2/3+2/6+2/12+....+2/x*(x+1) = 1+1989/1991
2 + 2/3 + 2/6 + 2/12 +...+ 2/x.(x+1) = 1+1989/1991
Giải đầy đủ hộ mình nhé!
