Question

Tìm xyz : x/4 =y/6=z/3 và xyz=57

Answer 1

Đặt x/4=y/6=z/3=k

=>x=4k; y=6k; z=3k

Ta có: xyz=57

=>72k³=57

hay \(k=\sqrt[3]{\dfrac{57}{72}}=\sqrt[3]{\dfrac{19}{24}}\)

=>\(x=4\sqrt[3]{\dfrac{19}{24}};y=6\sqrt[3]{\dfrac{19}{24}};z=3\sqrt[3]{\dfrac{19}{24}}\)

Answer 2

Áp dụng tính chất dãy tỷ số = :

\(\dfrac{x}{4}=\dfrac{y}{6}=\dfrac{z}{3}=\dfrac{x.y.z}{4.6.3}=\dfrac{57}{72}=\dfrac{19}{24}\\ =>\left\{{}\begin{matrix}x=\dfrac{19}{6}\\y=\dfrac{19}{4}\\z=\dfrac{19}{8}\end{matrix}\right.\)