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Ta có \(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2005}}\)
\(\Rightarrow\frac{1}{3}.B=\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2006}}\)
\(\Rightarrow B-\frac{1}{3}.B=\frac{1}{3}-\frac{1}{3^{2006}}\)
\(\frac{2}{3}.B=\frac{1}{3}-\frac{1}{3^{2006}}\)
\(B=\left(\frac{1}{3}-\frac{1}{3^{2006}}\right):\frac{2}{3}\)
\(B=\frac{1}{3}:\frac{2}{3}-\frac{1}{3^{2006}}:\frac{2}{3}=\frac{1}{2}-\frac{1}{2.3^{2005}}< \frac{1}{2}\)
Có B=\(\frac{1}{3}\)+\(\frac{1}{3^2}\)+\(\frac{1}{3^3}\)+...+\(\frac{1}{3^{2004}}\)+\(\frac{1}{3^{2005}}\)
=>3B=3.(\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\))
=>3B=1+\(\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\)
=>3B-B=(1+\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}\))-(\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2004}}+\frac{1}{3^{2005}}\))
=>2B=\(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+..+\frac{1}{3^{2003}}+\frac{1}{3^{2004}}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-....-\frac{1}{3^{2004}}-\frac{1}{3^{2005}}\)
=>2B=1-\(\frac{1}{3^{2005}}\)
=>B=(\(1-\frac{1}{3^{2005}}\)):2
Mà \(\left(1-\frac{1}{3^{2005}}\right)< \frac{1}{2}\)=>\(\left(1-\frac{1}{3^{2005}}\right):2< \frac{1}{2}\)
=>B<\(\frac{1}{2}\)(đpcm)
bạn ơi mình sửa cho bạn nè!
B=(1-\(\dfrac{1}{3^{2005}}\)) :2 = \(\dfrac{1}{2}\)-\(\dfrac{1}{\dfrac{3^{2005}}{2}}\) < \(\dfrac{1}{2}\)
1. a) \(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{1}{2}+\frac{1}{3}=\frac{9}{12}+\frac{6}{12}+\frac{4}{12}=\frac{19}{12}\)
b) \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)
\(=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}\)
\(=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}\)
\(=5+1+0,5=6,5\)
2) a) 1/2 + 2/3x = 1/4
=> 2/3x = 1/4 - 1/2
=> 2/3x = -1/4
=> x = -1/4 : 2/3
=> x = -3/8
b) 3/5 + 2/5 : x = 3 1/2
=> 3/5 + 2/5 : x = 7/2
=> 2/5 : x = 7/2 - 3/5
=> 2/5 : x = 29/10
=> x = 2/5 : 29/10
=> x = 4/29
c) x+4/2004 + x+3/2005 = x+2/2006 + x+1/2007
=> x+4/2004 + 1 + x+3/2005 + 1 = x+2/2006 + 1 + x+1/2007 + 1
=> x+2008/2004 + x+2008/2005 = x+2008/2006 + x+2008/2007
=> x+2008/2004 + x+2008/2005 - x+2008/2006 - x+2008/2007 = 0
=> (x+2008). (1/2004 + 1/2005 - 1/2006 - 1/2007) = 0
Vì 1/2004 + 1/2005 - 1/2006 - 1/2007 khác 0
Nên x + 2008 = 0 <=> x = -2008
Vậy x = -2008
1,a,\(\frac{3}{4}-\frac{-1}{2}+\frac{1}{3}=\frac{3}{4}+\frac{2}{4}+\frac{1}{3}=\frac{5}{4}+\frac{1}{3}=\frac{15}{12}+\frac{4}{12}=\frac{19}{12}\)
b, \(5\frac{5}{27}+\frac{7}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}=\frac{140}{27}-\frac{5}{27}+\frac{7}{23}+\frac{16}{23}+\frac{1}{2}=\frac{135}{27}+\frac{23}{23}+\frac{1}{2}=5+1+\frac{1}{2}=\frac{13}{2}\)2,a,\(\frac{1}{2}+\frac{2}{3}.x=\frac{1}{4}\)
<=>\(\frac{2}{3}.x=-\frac{1}{2}\)
<=>\(x=-\frac{3}{4}\)
b,\(\frac{3}{5}+\frac{2}{5}\div x=3\frac{1}{2}\)
<=>\(\frac{2}{5x}=\frac{29}{10}\)
<=>\(x=\frac{29}{4}\)
c,\(\frac{x+4}{2004}+\frac{x+3}{2005}=\frac{x+2}{2006}+\frac{x+1}{2007}\)
<=> \(\frac{x+4}{2004}+1+\frac{x+3}{2005}+1=\frac{x+2}{2006}+1+\frac{x+1}{2007}+1\)
<=>\(\frac{x+2008}{2004}+\frac{x+2008}{2005}=\frac{x+2008}{2006}+\frac{x+2008}{2007}\)
<=>\(\left(x+2008\right)\left(\frac{1}{2004}+\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\right)\)=0
<=>x+2008=0 vì cái ngoặc còn lại\(\ne0\)
<=>x=-2008
Vậy x=-2008
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