tìm x
1+ 1/3+1/6+1/10+...+1/x(x+2):2=1+ 2009/2011
Những câu hỏi liên quan
Tìm x biết 1+1/3+1/6+1/10+...+1/x(x+2):2=2009/2011
Tìm x€z, biết: 1/3+1/6+1/10+...+1/x(x+1):2=2009/2011
Tìm x:
1\3+1\6+1\10+...+2\x.(x+11) = 2009\2011
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(\frac{1}{x+1}=\frac{1}{2011}\)
\(x+1=2011\)
\(x=2010\)
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Tìm x biết : 1+1/3+1/6+1/10+...+1/x(x+1):2=1+2009/2011
1) tìm số tự nhiên x,biết rằng: 1/3+1/6+1/10+...+2/x(x+1)=2007/2009
2)so sánh:S=2/1*2*3+2/2*3*4+2/3*4*5+...+2/2009+2010+2011 và P=1/2
1.1/3+1/6+1/10+...+2/x.(x+1)=2007/2009
=>2/6+2/12+2/20+...+2/x.(x+1)=2007/2009
=>1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/(x+1)=2007/2009:2
=>1/2-1/(x+1)=2007/4018
=>1/(x+1)=1/2-2007/4018
=>1/x+1=1/2009
=>x+1=2009
=>x=2009-2008
=>x=1
vậy x=1
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làm đúng rồi nhưng phần:
x+1=2009
x=2009-1
x=2008
mà bạn
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Đặt A= \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2007}{2009}\)
\(\Rightarrow\) \(\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\times\left(x+1\right)}=\frac{2007}{4018}\)
\(\Leftrightarrow\frac{1}{2}A=\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{x\times\left(x+1\right)}=\frac{2007}{4018}\)
\(\Leftrightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2007}{4018}\)
\(\Leftrightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)
\(\Rightarrow A=1-\frac{2}{x+1}=\frac{2007}{2009}\)
\(\Rightarrow\frac{2}{x+1}=1-\frac{2007}{2009}\)
\(\Leftrightarrow\frac{2}{x+1}=\frac{2}{2009}\)
\(\Rightarrow x+1=2009\)
\(\Leftrightarrow x=2009-1\)
\(\Leftrightarrow x=2008\)
Vậy x=2008
1/3+1/6+1/10+...+2/x(x+1)=2009/2011
1/3+1/6+1/10 + ...+ 2/ x(x+1) = 2009/2011
hộ mk nha bạn nhanh 1h mk cần r
\(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4022}=\frac{1}{2011}\)
\(x+1=2011\)
\(x=2010\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2011}\)
\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x}+\frac{1}{x+1}\right)=\frac{2009}{2011}\)
phần sau tiếp tục nhé e
1/3 + 1/6 +1/10 +...+ 1/x.(x+1):2 = 2009/2011
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}...+\frac{1}{x\left(x+1\right):2}=\frac{2009}{2011}\)
\(=>\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2009}{4022}\)
\(=>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1009}{4022}\)
\(=>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(=>\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(=>\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4022}\)
\(=>\frac{1}{x+1}=\frac{1}{2011}\)
\(=>x+1=2011\)
\(=>x=2010\)
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\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{x.\left(x+1\right):2}=\frac{2009}{2011}\)
\(\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{x.\left(x+1\right):2}\right):2=\left(\frac{2009}{2011}\right):2\)
\(\Rightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{x.\left(x+1\right)}=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x.\left(x+1\right)}=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4022}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2011}\)
=> x + 1 = 2011
=> x = 2000
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Tìm x:
1/3+ 1/6+ 1/10+ ...+1/x(x+1)= 2009/2011
Nhân cả 2 vế với 1/2 ta có:
1/6+ 1/12+ 1/20+... +1/x(x+1)=2009/4022
1/2.3+ 1/3.4+ 1/4.5+... +1/x(x+1)=2009/4022
1/2- 1/3+ 1/3-1/4+ 1/4- 1/5+...+1/x- 1/x+1=2009/4022
1/2- 1/x+1=2009/4022
x+1/2.(x+1)- 2/2.(x+1)=2009/4022
x+1-2/2.(x+1)=2009/4022
x-1/2.(x+1)=2009/4022
=>(x-1). 4022= 2009.2.(x+1)
4022.x- 4022= 4018.x+ 4018
4022.x-4018.x= 4018+ 4022
x.(4022- 4018)= 8040
x.4 = 8040
x = 8040: 4
x =2010
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