Question

Tìm x,y,z

x³/8 = y³/27 = z³/64 và x² + 2y² - 3z² = -650

Answer 1

Ta có:

\(\dfrac{x^3}{8}=\dfrac{y^3}{27}=\dfrac{z^3}{64}=\dfrac{x}{\sqrt[3]{8}}=\dfrac{y}{\sqrt[3]{27}}=\dfrac{z}{\sqrt[3]{64}}=\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)và \(x^2+2y^2-3z^2=-650\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x^2}{2^2}=\dfrac{2y^2}{2.3^2}=\dfrac{3z^2}{3.4^2}=\dfrac{x^2+2y^2-3y^2}{4+18-48}=\dfrac{-650}{-26}=25\)

\(\dfrac{x}{2}=25\Rightarrow x=25.2=50\)

\(\dfrac{y}{3}=25\Rightarrow y=25.3=75\)

\(\dfrac{z}{4}=25\Rightarrow z=25.4=100\)

Vậy \(x=50;y=75;z=100\)