tim x : 1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)-1/1/2010
NN
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bai 11;tim x
1/(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)-1/x=1/2010
\(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2010}\).
\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}\)
\(=-\frac{1}{x+3}=\frac{1}{2010}\)
\(x=2010-\left(-3\right)=2013\)
tim x thuoc z biet
a,1/(1×2) + 1/(2×3) + ... + 2/(x(x+1)) = 2005/2010
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x.\left(x+1\right)}=\frac{2005}{2010}\)
\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{401}{402}\)
\(\Leftrightarrow1-\frac{1}{x+1}=\frac{401}{402}\)
\(\Leftrightarrow\frac{1}{x+1}=1-\frac{401}{402}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{402}\)
\(\Leftrightarrow x+1=402\Rightarrow x=401\)
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tim x
x+(x+1)+(x+2)+(x+3)+...+(x+2010)=2029099
Xét: \(1+2+3+.....+2010\) là dãy số tự nhiên cách đều
Tổng bằng:
\(\dfrac{\left(2010+1\right)\times\left[\left(2010-1\right):1+1\right]}{2}=\dfrac{2011\times2010}{2}=\dfrac{4042110}{2}=2021055\)(1)
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Tim x biet: 1/10+1/15+1/21+.........+2/x.(x+1) = 2010/2012
A= (1-1/2010) x (1- 2/2010) x (1- 3/2010) x.... x( 1- 2011/2010) =?
Vì ta có 1 - 1/2010 = 0/2010 = 0 nên suy ra biểu thức A = 0
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A=\(\left(1-\frac{1}{2010}\right).\left(1-\frac{2}{2010}\right)...\left(1-\frac{2010}{2010}\right)\left(1-\frac{2011}{2010}\right)\)
A=\(\frac{2009}{2010}.\frac{2008}{2010}...0.\frac{-1}{2010}\)
A=0
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tim x biet
x + 4/2009 + x + 3 / 2010 = x + 2/2011 + x + 1/2012
tim a va b de G(x)=x^2010+x^3+·^2+x+b chia het cho H(x)=X^2+x+1
tim so tu nhien x biat rang :1/10 +1/15+1/21+.....+2/x(x+1)=2010/2012
1/x×{x+1)+1/(x+1)×(x+2)+1/(x+2)×(x+3)-1/x=1/2010
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2010}\)
\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}\)
\(\frac{-1}{x+3}=\frac{1}{2010}\)
\(\Rightarrow-\left(x-3\right)=2010\)
\(\Rightarrow x=-2013\)
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\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2010}\)
\(\Rightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}\)
\(\Rightarrow\frac{1}{x}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}\)
\(\Rightarrow\left(\frac{1}{x}-\frac{1}{x}\right)-\frac{1}{x+3}=\frac{1}{2010}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{2010}\)
\(\Rightarrow x=2007\)
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\(\frac{1}{x.\left(x+1\right)}+\frac{1}{\left(x+1\right).\left(x+2\right)}+\frac{1}{\left(x+2\right).\left(x+3\right)}-\frac{1}{x}=\frac{1}{2010}\)
\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2010}\)
\(\frac{1}{x}-\frac{1}{x+3}-\frac{1}{x}=-\frac{1}{x+3}=\frac{1}{2010}\)
\(=>x=-2013\)
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