a) Áp dụng dãy tỉ số bằng nhau ta có:
$\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x+y+z}{2+3+5}=\dfrac{90}{10}=9$
$⇒\dfrac{x}{2}=9⇒x=18$
$\dfrac{y}{3}=9⇒x=27$
$\dfrac{z}{5}=9⇒x=45$
Vậy $(x,y,z)=(18,27,45)$
b) $2x=3y⇒\dfrac{x}{3}=\dfrac{y}{2}$
$3y=5z⇒\dfrac{z}{3}=\dfrac{y}{5}$
$⇒\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}$
mà $x-y+z=-33$
Áp dụng dãy tỉ số bằng nhau ta có:
$\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x-y+z}{15-10+6}=\dfrac{-33}{11}=-3$
$⇒\dfrac{x}{15}=-3⇒x=-45$
$\dfrac{y}{10}=-3⇒y=-30$
$\dfrac{z}{6}=-3⇒z=-18$
Vậy $(x,y,z)=(-45,-30,-18)$
c) $\dfrac{x}{3}=\dfrac{y}{4}$
$⇒x=\dfrac{y}{4}.3=\dfrac{3y}{4}$
$⇒xy=\dfrac{3y}{4}.y=\dfrac{3y^2}{4}=192$
$⇒3y^2=768$
$⇒y^2=758:3=256$
\(⇒\left[ \begin{array}{l}y=16\\x=12\end{array} \right.\)
\(⇒\left[ \begin{array}{l}y=-16\\x=-12\end{array} \right.\)
Vậy $(x,y)=(12,16),(-12,-16)$
