Ta có:
\(3^x+3^{-x}=3\)
\(\Leftrightarrow3^x\cdot\left(1+3^{-1}\right)=3\)
\(\Leftrightarrow3^x\cdot\left(1+\dfrac{1}{3}\right)=3\)
\(\Leftrightarrow3^x\cdot\dfrac{4}{3}=3\)
\(\Leftrightarrow3^x=3:\dfrac{4}{3}=\dfrac{9}{4}\)
\(\Leftrightarrow x=log_3\dfrac{9}{4}\)
a) Thay x vào A ta có:
\(A=3^{\dfrac{log_3\dfrac{9}{4}}{2}}+3^{-\dfrac{log_3\dfrac{9}{4}}{2}}=\dfrac{13}{6}\)
b) Thay x vào B ta có:
\(B=3^{2\cdot log_3\dfrac{9}{4}}+3^{-2\cdot log_3\dfrac{9}{4}}=\dfrac{6817}{1296}\)
Hiển nhiên \(A>0\)
\(A^2=\left(3^{\dfrac{x}{2}}+3^{-\dfrac{x}{2}}\right)^2=3^x+3^{-x}+2.3^{\dfrac{x}{2}-\dfrac{x}{3}}=3^x+3^{-x}+2=5\)
\(\Rightarrow A=\sqrt{5}\)
b.
\(9=\left(3^x+3^{-x}\right)^2=3^{2x}+3^{-2x}+2.3^{x-x}=3^{2x}+3^{-2x}+2\)
\(\Rightarrow3^{2x}+3^{-2x}=7\Rightarrow B=7\)
