\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{x.\left(x+1\right)}\)
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Ta thấy :
1/1x2 = 1/1 - 1/2
1/2x3 = 1/2 - 1/3
....
=>( 1/1x2 + 1/2x3 + 1/3x4 + 1/5x6 ) x 10 - x = ( 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 ) x 10 - x
= ( 1/1 - 1/6 ) x 10 - x =0
5/6 x 10 - x = 0
25/3 - x = 0
=> x = 25/3
1-1/x+1=2015/2016
=>1/x+1=1-2015/2016=1/2016
=>x+1=2016=>x=2015
mình không ghi lại đề nha:
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)
<=>\(1-\frac{1}{x+1}=\frac{2015}{2016}\)
<=>\(\frac{x}{x+1}=\frac{2015}{2016}\)
=>x=
Đến đó bạn tự giải tiếp ha
( 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 +1/5.6 ) x 10 - x = 0
= ( 1- 1/2 +1/2 -1/3 +1/3 - 1/4 + 1/4 - 1/5 +1/5 -1/6 ) x 10 - x = 0
= ( 1 - 1/6 ) x 10 - x = 0
= 5/6 x 10 - x =0
= 25/3 - x =0
x = 25/3 - 0
x = 25/3
\(\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\right)\times10-x=0\)
\(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\times10-x=0\)
\(\left(\frac{1}{1}-\frac{1}{6}\right)\times10-x=0\)
\(\frac{5}{6}\times10-x=0\)
\(\frac{25}{3}-x=0\)
x =\(\frac{25}{3}-0=\frac{25}{3}\)
\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{2009\cdot2010}\right)\cdot x=2009\)
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\right)\cdot x=2009\)
\(\left(1-\frac{1}{2010}\right)\cdot x=2009\)
\(\frac{2009}{2010}\cdot x=2009\)
\(x=2009:\frac{2009}{2010}\)
\(x=2010\)
\(\frac{1}{8}=12,5\%\) ; \(\frac{1}{16}=6,25\%\) ; \(\frac{1}{2}=50\%\) ; \(\frac{1}{4}=25\%\)
Thay vào trên mà tính.
= \(1+\left(\frac{3\left(1x2+2x4x2\right)}{3\left(5+5x3x25\right)}+1\right)-\left(1+\frac{18}{54}\right)-1\) = \(\frac{18}{380}-\frac{18}{54}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+..+\frac{1}{x\left(x+1\right)}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)
\(=1-\frac{1}{x+1}\)
\(=\frac{x+1-1}{x+1}=\frac{x}{x+1}\)
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{x.\left(x+1\right)}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)
\(=1-\frac{1}{x+1}\)
\(=\frac{x+1}{x+1}-\frac{1}{x+1}\)
\(=\frac{x}{x+1}\)