1+\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x.\left(x+1\right)}=1\frac{2008}{2010}\)
Những câu hỏi liên quan
a, \(\dfrac{5}{2}-3\left(\dfrac{1}{3}-x\right)=\dfrac{1}{4}-7x\)
b, \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2008}\right).x=\dfrac{2009}{1}+\dfrac{2010}{2}+...+\dfrac{5016}{2008}-2008\)
c, \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2001}{2003}\)
GIÚP VỚI , MIK CẦN GẤP
a)\(\frac{5}{2}-3\left(\frac{1}{3}-x\right)=\frac{1}{4}-7x\)
\(\Leftrightarrow\frac{5}{2}-1+x=\frac{1}{4}-7x\)
\(\Leftrightarrow8x=-\frac{5}{4}\)
\(\Leftrightarrow x=-\frac{5}{32}\)
c)\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2003}\)
\(\Leftrightarrow x+1=2003\)
\(\Leftrightarrow x=2002\)
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Tìm x, biết:
\(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}\right).x=\frac{2009}{1}+\frac{2010}{2}+\frac{2011}{3}+...+\frac{4016}{2008}-2008\)
Bài 2: Tìm x: (2đ)
b; \(\frac{1}{1} +\frac{1}{3}+\frac{1}{6} +...+\frac{2}{x.\left(x+1\right)}=1\frac{2008}{2010}\)
1.Tính tổngSleft(frac{-1}{7}right)^0+left(frac{-1}{7}right)^1+left(frac{-1}{7}right)^2+...+left(frac{-1}{7}right)^{2007}2.Tìm x5^x+5^{x+2}6503.CMRfrac{1}{6} frac{1}{5^2}+frac{1}{6^2}+frac{1}{7^2}+...+frac{1}{100^2} frac{1}{4}4. Cho Afrac{1}{2010}+frac{2}{2009}+frac{3}{2008}+...+frac{2009}{2}+frac{2010}{1} Bfrac{1}{2}+frac{1}{3}+frac{1}{4}+...+frac{1}{2010}+frac{1}{2011}So sánh A và B
Đọc tiếp
1.Tính tổng
\(S=\left(\frac{-1}{7}\right)^0+\left(\frac{-1}{7}\right)^1+\left(\frac{-1}{7}\right)^2+...+\left(\frac{-1}{7}\right)^{2007}\)
2.Tìm x
\(5^x+5^{x+2}=650\)
3.CMR
\(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)
4. Cho \(A=\frac{1}{2010}+\frac{2}{2009}+\frac{3}{2008}+...+\frac{2009}{2}+\frac{2010}{1}\)
\(B=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2010}+\frac{1}{2011}\)
So sánh A và B
giải phương trình: \(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}+...+\frac{x+2009}{2}+\frac{x+2010}{1}\)\(=\left(-2010\right)\)
\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}+...+\frac{x+2010}{1}=\left(-2010\right)\)
\(\Rightarrow\left(\frac{x+1}{2010}+1\right)+\left(\frac{x+2}{2009}+1\right)+...+\left(\frac{x+2010}{1}+1\right)=-2010+2010\)
\(\Rightarrow\frac{x+2011}{2010}+\frac{x+2011}{2009}+...+\frac{x+2011}{1}=0\)
\(\Rightarrow\left(x+2011\right)\left(1+\frac{1}{2}+...+\frac{1}{2009}+\frac{1}{2010}\right)=0\)
\(\Rightarrow x+2011=0\Leftrightarrow x=-2011\)
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cách bạn giải bài này là gì vậy ?
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CMR: \(\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-...+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}< 0,2\)
tìm x biết:
\(|x-\frac{1}{3}|=|\left(-3,2\right)+\frac{2}{5}|\)
\(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}\)
\(|x-\frac{1}{3}|=|\left(-3.2\right)+\frac{2}{5}|\)
\(\Rightarrow|x-\frac{1}{3}|=|-3.2+0.4|\)
\(\Rightarrow|x-\frac{1}{3}|=|-2.8|\)
\(\Rightarrow|x-\frac{1}{3}|=2.8\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=2.8\\x-\frac{1}{3}=-2.8\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{43}{15}\\x=-\frac{41}{15}\end{cases}}\)
tính lại kết quả nhé
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\(\left|x-\frac{1}{3}\right|=\left|-3.2+\frac{2}{5}\right|=\left|-\frac{14}{5}\right|\)\(=\frac{14}{5}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{14}{5}\\x-\frac{1}{3}=-\frac{14}{5}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\frac{47}{15}\\x=-\frac{37}{15}\end{cases}}\)
Vậy............
b,
\(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}-\frac{x-4}{2008}=0\)
\(\Leftrightarrow\frac{x-1}{2011}-1+\frac{x-2}{2010}-1-\frac{x-3}{2009}+1-\frac{x-4}{2008}+1=0\)
\(\Leftrightarrow\frac{x-1-2011}{2011}+\frac{x-2-2010}{2010}-\frac{x-3-2009}{2009}-\frac{x-4-2008}{2008}=0\)
\(\Leftrightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
\(\Leftrightarrow\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
Mà \(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)
=> x-2012=0
=>x=2012
Vậy ..............
TK MK NHA
*****CHUC BN HOC GIỎI*****
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Xem thêm câu trả lời
Tìm x thoả mãn:a)frac{1}{2}x-frac{3}{4}x-frac{7}{3}-frac{5}{6}b)frac{4}{5}x-x-frac{3}{2}x+frac{4}{3}frac{1}{2}-frac{6}{5}c)frac{1}{1.2}+frac{1}{2.3}+......+frac{1}{x.left(x+1right)}frac{2009}{2010}d)frac{1}{3}+frac{1}{6}+frac{1}{10}+...+frac{2}{xleft(x+1right)}1frac{2013}{2015}e)frac{1}{3.5}+frac{1}{5.7}+.....+frac{1}{left(2x+1right)left(2x+3right)}frac{100}{609}
Đọc tiếp
Tìm x thoả mãn:
a)\(\frac{1}{2}x-\frac{3}{4}x-\frac{7}{3}=-\frac{5}{6}\)
b)\(\frac{4}{5}x-x-\frac{3}{2}x+\frac{4}{3}=\frac{1}{2}-\frac{6}{5}\)
c)\(\frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{x.\left(x+1\right)}=\frac{2009}{2010}\)
d)\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2013}{2015}\)
e)\(\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{100}{609}\)
mk làm câu c cho nó dễ
c)1/1.2+1/2.3+...+1/x.(x+1)=2009/2010
=1-1/2+1/2-1/3+...+1/x-1/x+1=2009/2010
=1-1/x+1=2009/2010
=1/x+1=1-2009/2010
=1/x+1=1/2010
=) x+1=2010
x =2010-1
x =2009
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Đề cho dài :v. Lần sau đăng từ từ nhé bạn, hôm qua đến giờ mình giải không hết đó =(((
a) \(\frac{1}{2}.x-\frac{3}{4}.x-\frac{7}{3}=-\frac{5}{6}=\frac{-5}{6}\)
\(\frac{1}{2}.x-\frac{3}{4}.x=\frac{-5}{6}+\frac{7}{3}=\frac{3}{2}\)
\(\Leftrightarrow x\left(\frac{1}{2}-\frac{3}{4}\right)=\frac{3}{2}\Leftrightarrow x.\frac{-1}{4}=\frac{3}{2}\)
\(x=\frac{3}{2}:\frac{-1}{4}=-6\)
b) \(\frac{4}{5}.x-x-\frac{3}{2}.x+\frac{4}{3}=\frac{1}{2}-\frac{6}{5}=-\frac{7}{10}\)
\(\Leftrightarrow x\left(\frac{4}{5}-\frac{3}{2}.\frac{4}{3}\right)=x\left(\frac{4}{5}-2\right)=-\frac{7}{10}\)
\(\Leftrightarrow x.\frac{-6}{5}=-\frac{7}{10}\)
\(x=-\frac{7}{10}:\frac{-6}{5}=\frac{7}{12}\)
c) \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{2009}{2010}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{2010}\)
\(=1-\frac{1}{x+1}=\frac{2009}{2010}\)
\(\frac{1}{x+1}=1-\frac{2009}{2010}=\frac{1}{2010}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2010-1}=\frac{1}{2009}\). Vậy x= 2009
d) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=1\frac{2013}{2015}=\frac{4023}{2015}\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{4023}{2015}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{4023}{2015}\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{4023}{2015}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{4023}{2015}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{4023}{2015}:2=\frac{4023}{4030}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{4023}{4030}=\frac{-1004}{2015}=\frac{1004}{-2015}\)
\(x+1=\hept{\begin{cases}2015\\-2015\end{cases}}\Rightarrow x=\hept{\begin{cases}2014\\-2016\end{cases}}\)
e) Bạn tự làm, nhiều quá mình làm không hết
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tìm xa) frac{x-1}{2}+frac{x-2}{5}frac{1}{4}+frac{x-7}{10}b) 3-frac{2}{2x-3}frac{2}{5}+frac{1}{2x-3}-frac{3}{2}c)7cdotleft(x-1right)+2xcdotleft(1-xright)0d) frac{x+1}{2008}+frac{x+2}{2017}+frac{x+3}{2016}frac{x+10}{2009}+frac{x+11}{2008}+frac{x+12}{2007}e) frac{2}{left(x-1right)cdotleft(x-3right)}+frac{5}{left(x-3right)cdotleft(x-8right)}+frac{12}{left(x-8right)cdotleft(x-20right)}-frac{1}{x-20}frac{-3}{4}
Đọc tiếp
tìm x
a) \(\frac{x-1}{2}+\frac{x-2}{5}=\frac{1}{4}+\frac{x-7}{10}\)
b) \(3-\frac{2}{2x-3}=\frac{2}{5}+\frac{1}{2x-3}-\frac{3}{2}\)
c)\(7\cdot\left(x-1\right)+2x\cdot\left(1-x\right)=0\)
d) \(\frac{x+1}{2008}+\frac{x+2}{2017}+\frac{x+3}{2016}=\frac{x+10}{2009}+\frac{x+11}{2008}+\frac{x+12}{2007}\)
e) \(\frac{2}{\left(x-1\right)\cdot\left(x-3\right)}+\frac{5}{\left(x-3\right)\cdot\left(x-8\right)}+\frac{12}{\left(x-8\right)\cdot\left(x-20\right)}-\frac{1}{x-20}=\frac{-3}{4}\)
Tìm x: a,\(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}\)
b,\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{\left(2x-1\right)\left(2x+1\right)}=\frac{49}{99}\)
a, Ta có \(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}\)
<=> \(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}-\frac{x-4}{2008}=0\)
<=> \(\left(\frac{x-1}{2011}-1\right)+\left(\frac{x-2}{2010}-1\right)-\left(\frac{x-3}{2009}-1\right)-\left(\frac{x-4}{2008}-1\right)=0\)
<=>\(\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
<=> \(\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
Mà \(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)
=> \(x-2012=0=>x=2012\)
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b, \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{\left(2x-1\right)\left(2x+1\right)}=\frac{49}{99}\)
=>\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2x-1\right)\left(2x+1\right)}=2\cdot\frac{49}{99}\)
=>\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2x-1}-\frac{1}{2x+1}=\frac{98}{99}\)
=>\(1-\frac{1}{2x+1}=\frac{98}{99}\)
=>\(\frac{2x}{2x+1}=\frac{98}{99}\)
=>2x = 98
=>x = 49
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