Question

[Phạm Thanh Huyền]

Tìm x,y,z biết: \(\frac{3x}{8}=\frac{3y}{64}=\frac{3z}{216}\) và 2x² + 2y² - z² = 1

Answer 1

[Phạm Thanh Huyền]

Giải chi tiết hộ mình nha♥ Cám ơn các bạn !

Answer 2

[KAl(SO4)2·12H2O]

Em làm như sau nhé ;)

Ta có: \(\frac{3x}{8}=\frac{3y}{64}=\frac{3z}{216}\)

\(\Rightarrow\frac{x}{8}=\frac{y}{64}=\frac{z}{216}\Rightarrow\frac{x^2}{\left(8\right)^2}=\frac{y^2}{\left(64\right)^2}=\frac{z^2}{\left(216\right)^2}\)

\(\Rightarrow\frac{2x^2}{2.8^2}=\frac{2y^2}{2.64^2}=\frac{z^2}{216^2}\)

\(\Leftrightarrow\frac{2x^2+2y^2-z^2}{2.8^2+2.64^2-216^2}=\frac{1}{-38336}\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{8}=\frac{1}{-38336}\Rightarrow x=-4792\\\frac{y}{64}=\frac{-1}{-38336}\Rightarrow y=-599\\\frac{z}{216}=\frac{-1}{38336}\Rightarrow z=-\frac{4792}{27}\end{cases}}\)

\(\Rightarrow\left(x;y\right)\in\left\{-4792;-599;-\frac{4792}{27}\right\}\)