\(\dfrac{x}{3}\) = \(\dfrac{y}{4}\) = \(\dfrac{z}{5}\) và 2x2 + 2y2 + 3z2 = -100
2x = \(\dfrac{y}{3}\) = \(\dfrac{z}{5}\) và x + y - \(\dfrac{z}{2}\) = -20
\(\dfrac{15}{x-9}\) = \(\dfrac{20}{y-12}\) \(\dfrac{40}{z-24}\) và xy = 1200
\(\dfrac{40}{x-30}\) = \(\dfrac{28}{y-15}\) = \(\dfrac{28}{z-21}\) và xyz =...
Đọc tiếp
\(\dfrac{x}{3}\) = \(\dfrac{y}{4}\) = \(\dfrac{z}{5}\) và 2x2 + 2y2 + 3z2 = -100
2x = \(\dfrac{y}{3}\) = \(\dfrac{z}{5}\) và x + y - \(\dfrac{z}{2}\) = -20
\(\dfrac{15}{x-9}\) = \(\dfrac{20}{y-12}\) \(\dfrac{40}{z-24}\) và xy = 1200
\(\dfrac{40}{x-30}\) = \(\dfrac{28}{y-15}\) = \(\dfrac{28}{z-21}\) và xyz = 22400
a: Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\)
=>x=3k; y=4k; z=5k
\(2x^2+2y^2+3z^2=-100\)
=>\(2\left(3k\right)^2+2\cdot\left(4k\right)^2+3\cdot\left(5k\right)^2=-100\)
=>\(125k^2=-100\)
=>\(k^2=-\dfrac{4}{5}\)(vô lý)
vậy: \(\left(x;y;z\right)\in\varnothing\)
b: 2x=y/3=z/5
=>\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{3}=\dfrac{z}{5}=k\)
=>\(x=\dfrac{1}{2}k;y=3k;z=5k\)
\(x+y-\dfrac{z}{2}=-20\)
=>\(\dfrac{1}{2}k+3k-\dfrac{5k}{2}=-20\)
=>k=-20
=>\(x=\dfrac{1}{2}\cdot\left(-20\right)=-10;y=3\cdot\left(-20\right)=-60;z=5\cdot\left(-20\right)=-100\)