月份: 2021 年 2 月

(1) \(x^2-6x+9=0\)

(2) \(x^2+5x+3=0\)

(3) \(x^2-5x+4=0\)

(4) \(2x^2-6x+1=0\)

(5) \(4x^2-2x-1=0\)

(6) \(-2x^2-x-1=0\)

(7) \(5x^2-x-1=0\)

求証﹕\(\log_2(2^x+2^y)\geq\frac{x+y}{2}+1\).

証明﹕由AM-GM不等式得\(2^x+2^y\geq 2\sqrt{2^x\cdot 2^y}=2\sqrt{2^{x+y}}\)

\(\because y=\log_2x\)是增函數

\(\therefore \log_2(2^x+2^y)\geq \log_2{2\sqrt{2^{x+y}}}=\log_2 2+\log_2 2^{\frac{x+y}{2}}=1+\frac{x+y}{2}\)

証畢