Ҳисоб кунед:

\(\frac{a+b}{a-b}\), агар \(a^2+b^2=3ab\) ва 0 \(\lt b\lt a\) бошад.

Бигзор a=kb, дар ин ҷо k>1.

\(a^2+b^2=3ab\)

\(\frac{1}{b^2}\cdot(a^2+b^2)=\frac{1}{b^2}\cdot3ab\)

\(\frac{a^2}{b^2}+1=3\cdot{a}{b}\)

\(k=\frac{a}{b}\)

\(k^2+1=3\cdot k\)

\(k^2-3k+1=0\)

\(D=9-4=5\)

\(k_1=\frac{3+\sqrt{5}}{2}\)

\(k_2=\frac{3-\sqrt{5}}{2}\)

Азбаски k>1, \(k=\frac{3+\sqrt{5}}{2}\)

\(\frac{a+b}{a-b}=\frac{\frac{1}{b}\cdot(a+b)}{\frac{1}{b}\cdot(a-b)}=\)

\(=\frac{\frac{a}{b}+1}{\frac{a}{b}-1}=\)

\(=\frac{k+1}{k-1}=\frac{\frac{3+\sqrt{5}}{2}+1}{\frac{3+\sqrt{5}}{2}-1}=\)

\(=\frac{5+\sqrt{5}}{1+\sqrt{5}}=\frac{\sqrt{5}\cdot(\sqrt{5}+1}{1+\sqrt{5}}=\)

\(=\sqrt{5}\)

\(\frac{a+b}{a-b}=\sqrt{5}\)

Ҷавоб: \(\sqrt{5}\)