Question

Tìm x,y biết

\(\dfrac{x}{3}\)=\(\dfrac{y}{2}\) và \(3x^2\)- \(5y^2\)=-20

Answer 1

Ta có:

\(\dfrac{x}{3}=\dfrac{y}{2}\Leftrightarrow\dfrac{x^2}{9}=\dfrac{y^2}{4}\Leftrightarrow\dfrac{3x^2}{27}=\dfrac{5y^2}{20}\) và \(3x^2-5y^2=-20\)

áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{3x^2}{27}=\dfrac{5y^2}{20}=\dfrac{3x^2-5y^2}{27-20}=\dfrac{-20}{7}\)

+) \(\dfrac{3x^2}{27}=\dfrac{x^2}{9}=\dfrac{-20}{7}\Rightarrow7x^2=-20.9\Rightarrow7x^2=-180\Rightarrow x^2=-\dfrac{180}{7}\Rightarrow x=\sqrt{-\dfrac{180}{7}}\)+) \(\dfrac{5y^2}{20}=\dfrac{y^2}{4}=\dfrac{-20}{7}\Rightarrow7y^2=-20.4\Rightarrow7y^2=-80\Rightarrow y^2=\dfrac{-80}{7}\Rightarrow y=\sqrt{-\dfrac{80}{7}}\)Vậy ....