\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{x.\left(x+1\right)}\)
QA
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Cho B= \(\frac{1\times2}{1\times2\times3}+\frac{1\times2}{1\times2\times4}+\frac{1\times2}{1\times2\times3\times4}+\frac{1\times2}{1\times2\times3\times4\times5}+....+\frac{1\times2}{n,giao}\left(n\in N,n\ge3\right)\)
chứng tỏ B nhỏ hơn 3
\(\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{5\times6}\right)\times10-x=0\)
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
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\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{5.6}\right).10-x=0\)
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{30}\right).10-x=0\)
\(\left(1-\frac{1}{4}+\frac{1}{30}\right).10-x=0\)
\(\left(\frac{3}{4}+\frac{1}{30}\right).10-x=0\)
\(\frac{47}{60}.10-x=0\)
\(\frac{47}{6}-x=0\)
\(x=\frac{47}{6}-0\)
\(x=\frac{47}{6}\)
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Tìm x:
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{x\times\left(x+1\right)}=\frac{2015}{2016}\)
1-1/x+1=2015/2016
=>1/x+1=1-2015/2016=1/2016
=>x+1=2016=>x=2015
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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
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=>(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/x-1/x+1)=2015/2016
=>1/1-1/x+1=2015/2016
=>x/x+1=2015/2016
=>x=2015
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Tìm x biết \(\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\right)\times10-x=0\)
( 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
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\(\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}\right)\times10-x=0\)
\(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\times10-x=0\)
\(\left(\frac{1}{1}-\frac{1}{6}\right)\times10-x=0\)
\(\frac{5}{6}\times10-x=0\)
\(\frac{25}{3}-x=0\)
x =\(\frac{25}{3}-0=\frac{25}{3}\)
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Tìm \(x\):
\(\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+.....\frac{1}{2009\times2010}\right)\times x=2009\)
\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{2009\cdot2010}\right)\cdot x=2009\)
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2009}-\frac{1}{2010}\right)\cdot x=2009\)
\(\left(1-\frac{1}{2010}\right)\cdot x=2009\)
\(\frac{2009}{2010}\cdot x=2009\)
\(x=2009:\frac{2009}{2010}\)
\(x=2010\)
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\(\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+....+\frac{1}{2009}-\frac{1}{2010}\right).x=2009\)
\(\left(\frac{1}{1}-\frac{1}{2010}\right).x=2009\)
\(\frac{2009}{2010}.x=2009\)
\(x=2009:\frac{2009}{2010}\)
\(x=2010\)
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\(A=\left[1-\left(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+......+\frac{1}{98\times99\times100}\right)\right]\times\frac{14851}{19800}\)
Tính nhanh : \(\frac{1}{8}\div12,5\%+\left(\frac{1\times2\times3+6\times4\times2}{5\times3\times1+5\times15\times25}+\frac{1}{2}\div50\%\right)-\left(\frac{1}{16}\div6,25\%+\frac{3+2+1+2+4+6}{1+3+5+25+15+5}\right)-\frac{1}{4}\div25\%\)
\(\frac{1}{8}=12,5\%\) ; \(\frac{1}{16}=6,25\%\) ; \(\frac{1}{2}=50\%\) ; \(\frac{1}{4}=25\%\)
Thay vào trên mà tính.
= \(1+\left(\frac{3\left(1x2+2x4x2\right)}{3\left(5+5x3x25\right)}+1\right)-\left(1+\frac{18}{54}\right)-1\) = \(\frac{18}{380}-\frac{18}{54}\)
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Tính C=\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+....+\frac{1}{n\times\left(n+1\right)\times\left(n+2\right)}\)
Bạn nào giúp mik nhớ viết cả cách giải cho mik nhé!!!!!!!!!!
a,A=\(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
b,B=\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{998\times999\times100}\)
c,C=\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+98\right)}{1\times98+2\times97+3\times96+...+98\times1}\)