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\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{2001^2}+\frac{1}{2002^2}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{2000.2001}+\frac{1}{2001.2002}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{2000}-\frac{1}{2001}+\frac{1}{2001}-\frac{1}{2002}\)
\(\Rightarrow A< 1-\frac{1}{2002}=\frac{2001}{2002}\left(đpcm\right)\)
1)\(\frac{-8}{5}+\frac{207207}{201201}\)
=\(\frac{-8}{5}+\frac{207}{201}\)
=\(\frac{-8}{5}+\frac{69}{67}\)
=\(\frac{-191}{335}\)
Ta có \(VT=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2001}-\frac{1}{2002}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2001}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2002}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}+\frac{1}{2002}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2002}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}+\frac{1}{2002}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1001}\right)\)
\(=\frac{1}{1002}+...\frac{1}{2002}=VP\)
Vậy...
\(1-\frac{1}{2}+\frac{1}{3}-...+\frac{1}{2001}-\frac{1}{2002}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2001}\right)\)\(-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2002}\right)\)
= \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}+\frac{1}{2002}\right)\)\(-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2002}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2002}\right)\)\(-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1001}\right)\)
\(=\frac{1}{1002}+\frac{1}{1003}+\frac{1}{1004}+...+\frac{1}{2002}\)
Ta có :
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Rightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
Mà \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)
\(\Rightarrow x+2004=0\)
\(\Rightarrow x=-2004\)
Vậy ...
=>x+4/2000+1+x+3/2001+1=x+2/2002+1+x+1/2003+1
=>x+2004/2000+x+2004/2001=x+2004/2002+x+2004/2003
=>(x+2004)(1/2000+1/2001-1/2002-1/2003)=0
=>x+2004=0
=>x=-2004
Ta có \(\frac{x+1}{2001}+\frac{x+2}{2002}+\frac{x+3}{2003}+\frac{x+4}{2004}=4\)
\(\Rightarrow\frac{x+1}{2001}+\frac{x+2}{2002}+\frac{x+3}{2003}+\frac{x+4}{2004}-4=0\)
\(\Rightarrow\frac{x+1}{2001}-1+\frac{x+2}{2002}-1+\frac{x+3}{2003}-1+\frac{x+4}{2004}-1=0\)
\(\Rightarrow\frac{x+1-2001}{2001}+\frac{x+2-2002}{2002}+\frac{x+3-2003}{2003}+\frac{x+4-2004}{2004}=0\)
\(\Rightarrow\frac{x-2000}{2001}+\frac{x-2000}{2002}+\frac{x-2000}{2003}+\frac{x-2000}{2004}=0\)
\(\Rightarrow\left(x-2000\right)\left(\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}+\frac{1}{2004}\right)=0\)
Ta thấy ngay \(\Rightarrow\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}+\frac{1}{2004}\ne0\)
\(\Rightarrow x-2000=0\Rightarrow x=2000.\)
\(\frac{x+1}{2001}+\frac{x+2}{2002}+\frac{x+3}{2003}+\frac{x+4}{2004}=4\)
\(\Leftrightarrow\left(\frac{x+1}{2001}-1\right)+\left(\frac{x+2}{2002}-1\right)+\left(\frac{x+3}{2003}-1\right)+\left(\frac{x+4}{2004}-1\right)=0\)
\(\Leftrightarrow\left(\frac{x+1-2001}{2001}\right)+\left(\frac{x+2-2002}{2002}\right)+\left(\frac{x+3-2003}{2003}\right)+\left(\frac{x+4-2004}{2004}\right)=0\)
\(\Leftrightarrow\frac{x-2000}{2001}+\frac{x-2000}{2002}+\frac{x-2000}{2003}+\frac{x-2000}{2004}=0\)
\(\Leftrightarrow\left(x-200\right)\left[\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}+\frac{1}{2004}\right]=0\)
\(\Leftrightarrow x-2000=0\)
\(\Leftrightarrow x=2000\)
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 \(x+2004=0\Leftrightarrow x=-2004\)