Question

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}và3x^2-2y^2+z^2=5\)

Answer 1

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)

⇒ \(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{z^2}{25}\)

⇒ \(\dfrac{3x^2}{27}=\dfrac{2y^2}{32}=\dfrac{z^2}{25}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{3x^2}{27}=\dfrac{2y^2}{32}=\dfrac{z^2}{25}=\dfrac{3x^2-2y^2+z^2}{27-32+25}=\dfrac{5}{20}=\dfrac{1}{4}\)

⇒ \(\left\{{}\begin{matrix}x=\dfrac{1}{4}.3=\dfrac{3}{4}\\y=\dfrac{1}{4}.4=1\\z=\dfrac{1}{4}.5=\dfrac{5}{4}\end{matrix}\right.\)