Ta có :\(a^5=a^2.a^3\)
=> \(a^5=\left(\frac{3+\sqrt{5}}{2}\right)^2\left(\frac{3+\sqrt{5}}{2}\right)^3\)
=> \(a^5=\frac{\left(3+\sqrt{5}\right)^2}{2^2}.\frac{\left(3+\sqrt{5}\right)^3}{2^3}\)
=> \(a^5=\frac{\left(3+\sqrt{5}\right)^2\left(3+\sqrt{5}\right)^3}{2^2.2^3}\)
=> \(a^5=\frac{\left(9+6\sqrt{5}+5\right)\left(27+27\sqrt{5}+45+5\sqrt{5}\right)}{32}\)
=> \(a^5=\frac{\left(14+6\sqrt{5}\right)\left(72+32\sqrt{5}\right)}{32}\)
=> \(a^5=\frac{1008+432\sqrt{5}+448\sqrt{5}+960}{32}\)
=> \(a^5=\frac{1968+880\sqrt{5}}{32}=\frac{16\left(123+55\sqrt{5}\right)}{32}\)
=> \(a^5=\frac{123+55\sqrt{5}}{2}\)
Ta có :\(b^5=b^2.b^3\)
=> \(b^5=\left(\frac{3-\sqrt{5}}{2}\right)^2\left(\frac{3-\sqrt{5}}{2}\right)^3\)
=> \(b^5=\frac{\left(3-\sqrt{5}\right)^2}{2^2}.\frac{\left(3-\sqrt{5}\right)^3}{2^3}\)
=> \(b^5=\frac{\left(3-\sqrt{5}\right)^2\left(3-\sqrt{5}\right)^3}{2^2.2^3}\)
=> \(b^5=\frac{\left(9-6\sqrt{5}+5\right)\left(27-27\sqrt{5}+45-5\sqrt{5}\right)}{32}\)
=> \(b^5=\frac{\left(14-6\sqrt{5}\right)\left(72-32\sqrt{5}\right)}{32}\)
=> \(b^5=\frac{1008-432\sqrt{5}-448\sqrt{5}+960}{32}\)
=> \(b^5=\frac{1968-880\sqrt{5}}{32}=\frac{16\left(123-55\sqrt{5}\right)}{32}\)
=> \(b^5=\frac{123-55\sqrt{5}}{2}\)
Thay \(a^5=\frac{123+55\sqrt{5}}{2}\)và \(b^5=\frac{123-55\sqrt{5}}{2}\) vào biểu thức P ta được :
\(P=\frac{1}{\frac{123+55\sqrt{5}}{2}}+\frac{1}{\frac{123-55\sqrt{5}}{2}}\)
=> \(P=\frac{2}{123+55\sqrt{5}}+\frac{2}{123-55\sqrt{5}}\)
=> \(P=\frac{2\left(123-55\sqrt{5}\right)}{\left(123+55\sqrt{5}\right)\left(123-55\sqrt{5}\right)}+\frac{2\left(123+55\sqrt{5}\right)}{\left(123+55\sqrt{5}\right)\left(123-55\sqrt{5}\right)}\)
=>\(P=\frac{2\left(123-55\sqrt{5}\right)}{4}+\frac{2\left(123+55\sqrt{5}\right)}{4}\)
=>\(P=\frac{2\left(123-55\sqrt{5}\right)+2\left(123+55\sqrt{5}\right)}{4}\)
=> \(P=\frac{2\left(123-55\sqrt{5}+123+55\sqrt{5}\right)}{4}\)
=>\(P=\frac{2\left(123+123\right)}{4}=\frac{246}{2}=123\)
Vậy P = 123 .
