1/1x2 +1/2x3 +1/3x4 + 1/4x5 + ... + 1/2016 x 2017
1/1x2+1/2x3+1/3x4+1/4x5+...+1/X nhân ( X + 1 ) = 2017/2018 làm ơn đó
(1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100))
(1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100
haizzz đáng tiếc tôi muốn ns là: ko bao f và đừng mong chờ OK
1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100
(1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100
Lên Qanda mà hỏi
(1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100))
1/(1x2)/(2x3)/(3x4)):(1/(2x3)/(3x4)/(4x5)):...(1/(97*98)/(98*99)/(99*100
1/1x2 + 1/2x3 +1/3x4 +1/4x5 +1/5x6
1/1x2 + 1/2x3 + 1/3x4 +1/4x5 +1/5x6
= 1 -1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6
= 1 - 1/6 = 5/6
tìm x biết
a, (1/1x2+1/2x3+1/5x4+...+1/99x100) X=1/1x2+2x3+3x4+...+98x99
b, X/1x3+X/3x5+X/5x7+...+X/2013x2015=4/2015
c, X+1/2015+X+2/2016=X+3/2017+X+4/2018
b) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2013.2015}\)
\(=\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)
\(=\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{2015-2013}{2013.2015}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{2015}\right)=\frac{1007}{2015}\)
Phương trình tương đương với:
\(\frac{1007X}{2015}=\frac{4}{2015}\Leftrightarrow X=\frac{4}{1007}\)
c) \(\frac{x+1}{2015}+\frac{x+2}{2016}=\frac{x+3}{2017}+\frac{x+4}{2018}\)
\(\Leftrightarrow\frac{x+1}{2015}-1+\frac{x+2}{2016}-1=\frac{x+3}{2017}-1+\frac{x+4}{2018}-1\)
\(\Leftrightarrow\frac{x-2014}{2015}+\frac{x-2014}{2016}=\frac{x-2014}{2017}+\frac{x-2014}{2018}\)
\(\Leftrightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)