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a) \(\frac{1}{2010}\)và \(\frac{-7}{19}\)
Ta có : \(\frac{1}{2010}>0>\frac{-7}{19}\)
\(\Rightarrow\frac{1}{2010}>\frac{-7}{19}\)
b)\(\frac{497}{-499}\)và \(\frac{-2345}{2341}\)
Ta có : \(\frac{497}{-499}< -1< \frac{-2345}{2341}\)
\(\Rightarrow\frac{497}{-499}>\frac{-2345}{2341}\)
c)\(\frac{2000}{2001}\)và \(\frac{2001}{2002}\)
Ta có : \(\frac{2000}{2001}=1-\frac{1}{2001};\frac{2001}{2002}=1-\frac{1}{2002}\)
mà \(\frac{1}{2001}>\frac{1}{2002}\Rightarrow1-\frac{1}{2001}< 1-\frac{1}{2002}\)
\(\Rightarrow\frac{2000}{2001}< \frac{2001}{2002}\)
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Rightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+4}{2001}=\frac{x+4}{2002}+\frac{x+4}{2003}\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
vì \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\Rightarrow x+2004=0\)
=>x=-2004
vậy x=-2004
\(-\frac{1}{2003.2002}-\frac{1}{2002.2001}-....-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2002.2003}\right)\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2002}-\frac{1}{2003}\right)\)
\(=-\left(1-\frac{1}{2003}\right)\)
\(=\frac{-2002}{2003}\)
\(\frac{-1}{2003.2002}-\frac{1}{2002.2001}-\frac{1}{2001.2000}-....-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=-\left(\frac{1}{2003.2002}+\frac{1}{2002.2001}+\frac{1}{2001.2000}+....+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{2000.2001}+\frac{1}{2001.2002}+\frac{1}{2002.2003}\right)\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{2001}-\frac{1}{2002}+\frac{1}{2002}-\frac{1}{2003}\right)\)
\(=-\left(1-\frac{1}{2003}\right)\)
\(=-\frac{2002}{2003}\)
tham khảo bài của mình tại http://olm.vn/hoi-dap/question/133172.html
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có: \(\frac{a+2001}{b+2001}=\frac{a}{b}=\frac{2001}{2001}=1\)
\(\Rightarrow\frac{a}{b}=\frac{a+2001}{b+2001}\)
ta xét tích
a( b +2001) = ab + 2001a
b(a + 2001) = ab + 2001b
vì b > 0 => b+ 2001>0
+) a>b => ab + 2001a > ab + 2001b
=> \(\frac{a}{b}>\frac{a+2001}{b+2001}\)
+) a < b => ab + 2001a < ab + 2001b
=> \(\frac{a}{b}< \frac{a+2001}{b+2001}\)
+) a = b
=> \(\frac{a}{b}=\frac{a+2001}{b+2001}\)
a,b=0;1
nếu a,b=0 thì:a^2011+b^2011=0+0=0
nếu a,b=1 thì:a^2011+b^2011=1+1=2
T*** mik nha!
TH1 a,b=0 kết quả tính=0
TH2 a,b=1 kết quả tính=2
Còn lại ko còn Th nào thảo mãn