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Ta có: \(A=\dfrac{3^{10}+1}{3^9+1}\)
\(\Leftrightarrow A=\dfrac{3^{10}+3-2}{3^9+1}\)
hay \(A=3-\dfrac{2}{3^9+1}\)
Ta có: \(B=\dfrac{3^9+1}{3^8+1}\)
\(\Leftrightarrow B=\dfrac{3^9+3-2}{3^8+1}\)
hay \(B=3-\dfrac{2}{3^8+1}\)
Ta có: \(3^9+1>3^8+1\)
\(\Leftrightarrow\dfrac{2}{3^9+1}< \dfrac{2}{3^8+1}\)
\(\Leftrightarrow-\dfrac{2}{3^9+1}>-\dfrac{2}{3^8+1}\)
\(\Leftrightarrow-\dfrac{2}{3^9+1}+3>-\dfrac{2}{3^8+1}+3\)
hay A>B
Ta có:
\(\frac{1}{3}\)A = \(\frac{3^{10}+1}{3^{10}+3}\)
= \(\frac{3^{10}+1}{3^{10}+1+2}\)
= \(1+\frac{3^{10}+1}{2}\)
\(\frac{1}{3}\)B = \(\frac{3^9+1}{3^9+3}\)
= \(\frac{3^9+1}{3^9+1+2}\)
= 1 + \(\frac{3^9+1}{2}\)
Đương nhiên \(1+\frac{3^{10}+1}{2}\) > 1 + \(\frac{3^9+1}{2}\)
=> A > B