Question

Tìm x, y biết 3x = 5y và \(3x^2-2y^2=3\)

Answer 1

\(3x=5y\Rightarrow\frac{x}{5}=\frac{y}{3}\).

Tính chất dãy tỉ bằng nhau \(\frac{x}{5}=\frac{y}{3}=\frac{x-y}{5-3}=\frac{3x^2-2y^2}{3.5^2-2.3^2}=\frac{3}{57}\)

Với \(3x=\frac{3}{57}\Rightarrow x=\frac{1}{57}\)Với \(5y=\frac{3}{57}\Rightarrow y=\frac{1}{95}\)

Answer 2

Từ \(3x=5y\Rightarrow\frac{x}{5}=\frac{y}{3}\Rightarrow\frac{x^2}{25}=\frac{y^2}{9}\Rightarrow\frac{3x^2}{75}=\frac{2y^2}{18}=\frac{3x^2-2y^2}{75-18}=\frac{3}{57}=\frac{1}{19}\)

\(\left(+\right)\frac{3x^2}{75}=\frac{1}{19}\Rightarrow\frac{x^2}{25}=\frac{1}{19}\Rightarrow x^2=\frac{25}{19}\Rightarrow x=\frac{5}{\sqrt{19}}\)

\(\left(+\right)\frac{2y^2}{18}=\frac{1}{19}\Rightarrow\frac{y^2}{9}=\frac{1}{19}\Rightarrow y^2=\frac{9}{19}\Rightarrow y=\frac{3}{\sqrt{19}}\)