2d:
Theo đề bài
→ $x=\dfrac{2}{3}y$, $z=\dfrac{5}{3}y$
→ $xyz=\dfrac{2}{3}y.y.\dfrac{5}{3}y$
→ $xyz=\dfrac{10}{9}y^3=810$
→ $y^3=729=9^3$
→ $y=9$
→ $x=\dfrac{2}{3}.9=6$
→ $z=\dfrac{5}{3}.9=15$
3
Ta có: $ABCD$ là hình tứ giác
→ $A+B+C+D=360^o$ mà $\dfrac{A}{2}=\dfrac{B}{3}=\dfrac{C}{4}=\dfrac{D}{6}$
→ $\dfrac{A}{2}=\dfrac{B}{3}=\dfrac{C}{4}=\dfrac{D}{6}=\dfrac{A+B+C+D}{2+3+4+6}=\dfrac{360^o}{15}=24$
→ $\dfrac{A}{2}=24$ → $A=48^o$
$\dfrac{B}{3}=24$ → $B=72^o$
$\dfrac{C}{4}=24$ → $C=96^o$
$\dfrac{D}{6}=24$ → $D=144^o$
