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\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x.\left(x-1\right)}=\frac{2007}{2009}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}\right)+2.\left(\frac{1}{3}-\frac{1}{4}\right)+....+2.\left(\frac{1}{x-1}-\frac{1}{x}\right)=\frac{2007}{2009}\)
\(2.\frac{1}{2}-2.\frac{1}{x}=\frac{2007}{2009}\)
\(\frac{1}{2}-\frac{1}{x}=\frac{2007}{4018}\)
\(\frac{1}{x}=\frac{1}{2}-\frac{2007}{4018}\)
\(\frac{1}{x}=\frac{1}{2009}\)
x=2009
22.3+23.4+...+2x.(x−1)=20072009
2.(12−13)+2.(13−14)+....+2.(1x−1−1x)=20072009
2.12−2.1x=20072009
12−1x=20074018
1x=12−20074018
1x=12009
x=2009
Kết quả :
1/3 + 1/6 + 1/10 + 1/15 + ...+2/X x (X + 1 )
= 2 x ( 1/6 + 1/12 +1/20 + 1/30 + ...+ 1/X x ( X + 1 )
= 2 x (1/ 2x3 + 1/3x4 + 1/4x5 + 1/5x6 +...+1/x - 1/x+1 )
= 2 x ( 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 -1/6 + ...+1/x - 1/x+1 )
= 2 x ( 1/2 - 1/x+1 ) = 2007/2009
= 1 - 2/x+ 1 = 2007/2009
2/x+1 = 1 - 2007/2009
2/x + 1= 2/2009
x + 1 = 2009
x = 2009 - 1
x = 2008
\(\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\) trong đó 3 = x ; 2 = x - 1
\(\frac{1}{\left(x-1\right)x}=\frac{1}{x-1}-\frac{1}{x}\)
ĐẶt A = \(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x-1\right)}-\frac{2007}{2009}\)
A = \(\frac{2}{6}+\frac{2}{12}+\frac{2}{30}\cdot\cdot\cdot+\frac{2}{x\left(x-1\right)}-\frac{2007}{2009}\)
A = \(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+..+\frac{2}{\left(x-1\right)x}-\frac{2007}{2009}\)
A = \(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x-1}-\frac{1}{x}-\frac{2007}{2009}\)
A \(=\frac{1}{2}-\frac{1}{x}-\frac{2007}{2009}\)
Bài 3 :
b) Ta có 1+ 2 + 3 +4 + ...+ x =15
Nên \(\frac{x\left(x+1\right)}{2}=15\)
\(x\left(x+1\right)=30\)
=> \(x\left(x+1\right)=5.6\)
=> x = 5
= 12009 nha mình đúng đấy mình nha
thanks