2 câu trả lời
Ta có
$\dfrac{x}{2} . \dfrac{x}{2} . \dfrac{x}{2} = \dfrac{x}{2} . \dfrac{y}{3} . \dfrac{z}{7}$
$\dfrac{x^3}{8} = \dfrac{xyz}{2.3.7} = \dfrac{42}{42}$
Vậy $x^3 = 8$, suy ra $x = 2$
Từ $\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z}{7}$, ta suy ra $y = 3$ và $z = 7$
$C1$ . $\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{x}{7}$
$⇔ \dfrac{x}{2} = \dfrac{y}{3} = \dfrac{x}{7} = \dfrac{x.y.z}{2.3.7}$
$⇒ \dfrac{42}{42} = 1$
$⇒$ $\left \{ {{x=2} \atop {y=3}} \atop {z=7} \right.$
Vậy `(x;y;z)=(2;3;7)`
$C2$. Đặt $\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z}{7} = k$
$⇒$ $x = 2k; y = 3k; z = 7k$
$⇒ x.y.z = 2k.3k.7k = 42$
$⇔ k^3 = 1$
$⇔ k=1$
Dễ dàng tìm được : $\left \{ {{x=2} \atop {y=3}} \atop {z=7} \right.$
Vậy `(x;y;z)=(2;3;7)`
