Tìm x, y, z biết \(\frac{x}{2}\,\, = \,\,\frac{y}{3}\,\, = \,\,\frac{z}{4}\) và 2x² + 3y² – 5z² = –405.

Ta có: \(\frac{x}{2}\,\, = \,\,\frac{y}{3}\,\, = \,\frac{z}{4}\)⇒ \(\frac{{{x^2}}}{4}\,\, = \,\,\frac{{{y^2}}}{9}\,\, = \,\,\frac{{{z^2}}}{{16}}\)⇒ \(\frac{{2{x^2}}}{8}\,\, = \,\,\frac{{3{y^2}}}{{27}}\,\, = \,\,\frac{{5{z^2}}}{{80}}\)

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\frac{{2{x^2}}}{8}\,\, = \,\,\frac{{3{y^2}}}{{27}}\,\, = \,\,\frac{{5{z^2}}}{{80}}\,\, = \,\,\frac{{2{x^2} + \,3{y^2}\, - \,5{z^2}\,}}{{8\, + \,27\, - \,80}}\,\, = \,\,\frac{{ - 405}}{{ - 45}}\,\, = \,\,9\).

⇒\(\left\{ \begin{array}{l}\frac{{{x^2}}}{4}\, = \,9\\\frac{{{y^2}}}{9}\, = \,9\\\frac{{{z^2}}}{{16}}\, = \,9\end{array} \right.\)   ⇒\(\left\{ \begin{array}{l}{x^2}\, = \,36\\{y^2}\, = \,81\\{z^2}\, = \,144\end{array} \right.\)   ⇒\(\left\{ \begin{array}{l}x\, = \, \pm 6\\y\, = \, \pm 9\\z\, = \, \pm 12\end{array} \right.\) .