3.
\(\sqrt{log_3 \sqrt{6}+\sqrt{log_3 2}}+\sqrt{log_3 \sqrt{6}-\sqrt{log_3 2}}=\)？
[解答]
下面\(A,B\)沒有定義的很嚴謹，看得懂就好
\(\sqrt{A}+\sqrt{B}=\sqrt{A+B+2\sqrt{AB}}\)
\(\displaystyle A+B=\log_3\sqrt{6}=\frac{1}{2}+\frac{1}{2}\log_32\)
\(\displaystyle AB=\frac{1}{4}\log_32\)
不難看出\(\displaystyle (A,B)=(\frac{1}{2},\frac{1}{2}\log_32)\)
(補)試了一下你說的也行
\(A^2+B^2=\log_36=1+\log_32=1+t\)
\(\displaystyle AB=\sqrt{(\frac{1}{2}(1+\log_32))^2-\log_32}\)
\(\displaystyle =\sqrt{\frac{1}{4}t^2-\frac{1}{2}t+\frac{1}{4}}=\frac{1}{2}(1-t)\)
\(A^2+2AB+B^2=2,A+B=\sqrt{2}\)
