Giải:
\(A=\dfrac{2005^2-2004}{2005^3+1}\)
\(\Leftrightarrow A=\dfrac{2005^2-2005+1}{\left(2005+1\right)\left(2005^2-2005+1\right)}\)
\(\Leftrightarrow A=\dfrac{1}{2005+1}\left(1\right)\)
\(B=\dfrac{2005^2+2006}{2005^3-1}\)
\(\Leftrightarrow B=\dfrac{2005^2+2005+1}{\left(2005-1\right)\left(2005^2+2005+1\right)}\)
\(\Leftrightarrow B=\dfrac{1}{2005-1}\left(2\right)\)
Ta có:
\(\left(1\right)< \left(2\right)\)
\(\Leftrightarrow A< B\)
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