Question

x:2=y:3=z:5 và x^2+y^2+z^2=152

Answer 1

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\) va \(x^2+y^2+z^2=152\)

Ta co: \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)

⇒ \(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{25}\) ⇒ \(\dfrac{x^2+y^2+z^2}{4+9+25}\) ⇒ \(\dfrac{x^2+y^2+z^2}{38}\)

Voi \(x^2+y^2+z^2=152\) ⇒ \(\dfrac{152}{38}=4\)

⇒ \(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{25}=4\)

⇒ \(\dfrac{x^2}{4}=4\) ⇒ \(x^2=16\) ⇒ \(x=\pm4\)

\(\dfrac{y^2}{9}=4\) ⇒ \(y^2=36\) ⇒ \(y=\pm6\)

\(\dfrac{z^2}{25}=4\) ⇒ \(z^2=100\) ⇒ \(z=\pm10\)

Vay \(x=\pm4\) ; \(y=\pm6\) ; \(z=\pm10\)

Answer 2

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\&x^2+y^2+z^2=152\)

Từ \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{25}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{25}=\dfrac{x^2+y^2+z^2}{4+9+25}=\dfrac{152}{38}=4\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x^2}{4}=4\\\dfrac{y^2}{9}=4\\\dfrac{z^2}{25}=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^2=16\\y^2=36\\z^2=100\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm4\\y=\pm6\\z=\pm10\end{matrix}\right.\)

Vậy, các cặp (x;y;z) thỏa mãn là: (4;6;10) và (-4;-6;-10)

Answer 3

Từ x:2 = y : 3 = z : 5 => \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\) = \(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{25}\)

=> \(\dfrac{x^2+y^2+z^2}{4+9+25}=\dfrac{152}{38}=4\)

\(\Rightarrow\dfrac{x^2}{4}=4;\dfrac{y^2}{9}=4;\dfrac{z^2}{25}=4\)

Với \(\dfrac{x^2}{4}=4\Rightarrow x^2=16\Rightarrow x=\pm4\)

Với \(\dfrac{y^2}{9}=4\Rightarrow y^2=36\Rightarrow y=\pm6\)

Với \(\dfrac{z^2}{25}=4\Rightarrow z^2=100\Rightarrow z=\pm10\)