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B. 1/3 - 1/3 - 3/5 +3/5 + 5/7 - 5/7 + 9/11 - 9/11 -11/13 + 11/ 13 + 7/9 + 13/15
= 0 -0-0-0-0+7/9 +13/15
= 74/45
2.
a) Ta có:
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)=\left(x+1\right)\left(\frac{1}{13}+\frac{1}{14}\right)\)
Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\ne\frac{1}{13}+\frac{1}{14}\)nên \(x+1=0\Leftrightarrow x=-1\)
Vậy x = -1
b) Ta có:
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Rightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}\right)=\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{2003}\right)\)
Vì \(\frac{1}{2000}+\frac{1}{2001}\ne\frac{1}{2002}+\frac{1}{2003}\)nên \(x+2004=0\Leftrightarrow x=-2004\)
Vậy, x = -2004
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.....+\frac{1}{2001^2}+\frac{1}{2002^2}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{2000.2001}+\frac{1}{2001.2002}\)
\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{2000}-\frac{1}{2001}+\frac{1}{2001}-\frac{1}{2002}\)
\(\Rightarrow A< 1-\frac{1}{2002}=\frac{2001}{2002}\left(đpcm\right)\)