Khi f₁  $$\to \cos {{\phi }_{1}}=1\to 2\pi {{f}_{1}}L=\frac{1}{2\pi {{f}_{1}}C}=a$$ 

Đặt f₂ = n.f₁   $$\to \cos {{\phi }_{2}}=0,6\to \tan {{\phi }_{2}}=\frac{4}{3}\leftrightarrow \frac{n.2\pi {{f}_{1}}L-\frac{1}{n.2\pi {{f}_{1}}C}}{R}=\frac{4}{3}\leftrightarrow \frac{n.a-\frac{a}{n}}{R}=\frac{4}{3}$$  (1)

Đặt f₃ = m.f₁  $$\to \cos {{\phi }_{3}}=\frac{15}{17}\to \tan {{\phi }_{3}}=\frac{8}{15}\leftrightarrow \frac{m.2\pi {{f}_{1}}L-\frac{1}{m.2\pi {{f}_{1}}C}}{R}=\frac{8}{15}\leftrightarrow \frac{m.a-\frac{a}{m}}{R}=\frac{8}{15}$$ (2)

Lấy (1) chia (2) ta có:  $$\frac{n.-\frac{1}{n}}{m.-\frac{1}{m}}=\frac{5}{2}\to 2n-\frac{2}{n}=5m-\frac{5}{m}$$ (3)

Vì: f₂ = n.f₁ f₁ + 150 = n.f₁ f₁ (n-1) = 150 (4)

Vì: f₃ = m.f₁ f₁ + 50 = m.f₁ f₁ (m-1) = 50(5)

Lấy (4) chia (5) ta có: $$\frac{n-1}{m-1}=3\to n=3m-2$$