Question

Giải pt : \(\sqrt{\left(9+4\sqrt{5}\right)^x}+\sqrt{\left(9-4\sqrt{5}\right)^x}=18\)

Answer 1

\(pt\Leftrightarrow\left(9+4\sqrt{5}\right)^{\dfrac{x}{2}}+\left(9-4\sqrt{5}\right)^{\dfrac{x}{2}}=18\)

Thấy rằng \(9-4\sqrt{5}\) là nghịch đảo của \(9+4\sqrt{5}\)

Do vậy \(\left(9+4\sqrt{5}\right)^{\dfrac{x}{2}}\left(9-4\sqrt{5}\right)^{\dfrac{x}{2}}=1\)

Đặt ​\(\left(9-4\sqrt{5}\right)^{\dfrac{x}{2}}=t\) ta có pt:​

\(t+\dfrac{1}{t}=18\Rightarrow t^2-18t+1=0\Rightarrow t=9\pm4\sqrt{5}\)

Vì vậy \(t=9\pm4\sqrt{5}=\left(9-4\sqrt{5}\right)^{\pm1}=\left(9-4\sqrt{5}\right)^{\dfrac{x}{2}}\)

\(\Rightarrow\dfrac{x}{2}=\pm1\Rightarrow x=\pm2\)

Answer 2

\(\sqrt{\left(9+4\sqrt{5}\right)^x}+\sqrt{\left(9-4\sqrt{5}\right)^x}=18\)

<=>\(\sqrt{\left(5+2.2\sqrt{5}+4\right)^x}+\sqrt{\left(5-2.2.\sqrt{5}+4\right)^x}=18\)

<=>\(\sqrt{\left(\sqrt{5}+2\right)^{2x}}+\sqrt{\left(\sqrt{5}-2\right)^{2x}}=18\)

<=>\(\left(\sqrt{5}+2\right)^x+\left(\sqrt{5}-2\right)^x=18\)

Nhận xét:

x>2 thì VT>18=VP

x<2 thì VT<18=VP

x=2 thì VT=VP

Vậy S={2}