Question

Tìm x;y;z biết \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{4}=\dfrac{z}{5}\)và \(x^2\) \(-y^2=-16\)

Answer 1

Ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}\Leftrightarrow\left(\dfrac{x}{2}\right)^2=\left(\dfrac{y}{3}\right)^2\Leftrightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{x^2-y^2}{4-9}=\dfrac{-16}{-5}=3,2\)

Do đó:

x² = \(4.3,2=12,8\Rightarrow x=\sqrt{12,8}\)

y² = \(9.3,2=28,8\Rightarrow y=\sqrt{28,8}\)

Suy ra: \(\dfrac{\sqrt{28,8}}{4}=\dfrac{z}{5}\Rightarrow z=\dfrac{5.\sqrt{28,8}}{4}=3\sqrt{5}\)

Vậy \(x=\sqrt{12,8};y=\sqrt{28,8};z=3\sqrt{5}\)