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\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{2001^2}+\frac{1}{2002^2}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{2000.2001}+\frac{1}{2001.2002}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{2000}-\frac{1}{2001}+\frac{1}{2001}-\frac{1}{2002}\)
\(\Rightarrow A< 1-\frac{1}{2002}=\frac{2001}{2002}\left(đpcm\right)\)
Giải bài khó nhất =)
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Leftrightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)=0\)
Do \(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\ne0\) nên \(x+2004=0\Leftrightarrow x=-2004\)
Đặt \(A=\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-...-\frac{1}{2^{2004}}\)
\(2^2A=1-\frac{1}{2^2}+\frac{1}{2^4}-...-\frac{1}{2^{2002}}\)
\(2^2A+A=\left(1-\frac{1}{2^2}+\frac{1}{2^4}-...-\frac{1}{2^{2002}}\right)+\left(\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-...-\frac{1}{2^{2004}}\right)\)
\(5A=1-\frac{1}{2^{2004}}\)
\(\Rightarrow5A< 1\Rightarrow A< \frac{1}{5}\left(đpcm\right)\)
Đặt A = \(\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-\frac{1}{2^8}+...+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}\)
=> 22A = 4A = \(1-\frac{1}{2^2}+\frac{1}{2^4}-\frac{1}{2^6}+...+\frac{1}{2^{2000}}-\frac{1}{2^{2002}}\)
=> 4A + A =\(1-\frac{1}{2^2}+\frac{1}{2^4}-\frac{1}{2^6}+...+\frac{1}{2^{2000}}-\frac{1}{2^{2002}}+\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-\frac{1}{2^8}+...+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}\)
=> 5A = \(1-\frac{1}{2^{2004}}\)
=> \(A=\frac{1}{5}-\frac{1}{2^{2004}.5}< \frac{1}{5}=0,2\)
=> A < 0,2 (ĐPCM)
Có \(\frac{x+4}{2000}\) + \(\frac{x+3}{2001}\) = \(\frac{x+2}{2002}\) + \(\frac{x+1}{2003}\)
( \(\frac{x+4}{2000}\) + 1 ) + ( \(\frac{x+3}{2001}\) + 1 ) = ( \(\frac{x+2}{2002}\) + 1 ) + ( \(\frac{x+1}{2003}\) + 1 )
( \(\frac{x+4}{2000}\) + \(\frac{2000}{2000}\) ) + ( \(\frac{x+3}{2001}\) + \(\frac{2001}{2001}\) ) = ( \(\frac{x+2}{2002}\) + \(\frac{2002}{2002}\) ) + ( \(\frac{x+1}{2003}\) + \(\frac{2003}{2003}\) )
\(\frac{x+4+2000}{2000}\) + \(\frac{x+3+2001}{2001}\) = \(\frac{x+2+2002}{2002}\) + \(\frac{x+1+2003}{2003}\)
\(\frac{x+2004}{2000}\) + \(\frac{x+2004}{2001}\) = \(\frac{x+2004}{2002}\) + \(\frac{x+2004}{2003}\)
\(\frac{x+2004}{2000}\) + \(\frac{x+2004}{2001}\) - \(\frac{x+2004}{2002}\) - \(\frac{x+2004}{2003}\) = 0
( x + 2004 ) + ( \(\frac{1}{2000}\) + \(\frac{1}{2001}\) + \(\frac{1}{2002}\) + \(\frac{1}{2003}\) ) = 0
Mà \(\frac{1}{2000}\) + \(\frac{1}{2001}\) + \(\frac{1}{2002}\) + \(\frac{1}{2003}\) \(\ne\) 0
\(\Rightarrow\) x + 2004 = 0
\(\Rightarrow\) x = -2004
Vậy x = - 2014
\(2A=1+\frac{1}{2}+...+\frac{1}{2^{49}}\)
\(2A-A=1-\frac{1}{2^{50}}\)
\(A=1-\frac{1}{2^{50}}\)=> A bé hơn 1
tương tự nha
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)
\(2A=2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(A=1-\frac{1}{2^{50}}< 1\)
2: =>2x-1/4=5/6-1/2x
=>5/2x=5/6+1/4=13/12
=>x=13/30
3: =>3x-5/6=2/3-1/2x
=>3,5x=2/3+5/6=4/6+5/6=9/6=3,2
hay x=32/35
1)\(\frac{-8}{5}+\frac{207207}{201201}\)
=\(\frac{-8}{5}+\frac{207}{201}\)
=\(\frac{-8}{5}+\frac{69}{67}\)
=\(\frac{-191}{335}\)