Đặt \(\frac{x}{4}=\frac{y}{3}=\frac{z}{5}=kak\left(kak\ne0\right)\)
\(\Rightarrow\hept{\begin{cases}x=4kak\\y=3kak\\z=5kak\end{cases}}\)
Mà \(x^2+y^2+z^2=200\)
\(\Leftrightarrow\left(4kak\right)^2+\left(3kak\right)^2+\left(5kak\right)^2=200\)
\(\Leftrightarrow16.kak^2+9.kak^2+25.kak^2=200\)
\(\Leftrightarrow kak^2.\left(16+9+25\right)=200\)
\(\Leftrightarrow kak^2.50=200\)
\(\Leftrightarrow kak^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}kak=2\\kak=-2\end{cases}}\)
+) Với \(kak=2\)thì \(\hept{\begin{cases}x=4kak=8\\y=3kak=6\\z=5kak=10\end{cases}}\)
+) Với \(kak=-2\)thì \(\hept{\begin{cases}x=4kak=-8\\y=3kak=-6\\z=5kak=-10\end{cases}}\)
Vậy ...
Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\left(k\ne0\right)\)
\(\Rightarrow\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)
Ta có : \(xyz=-30\)
\(\Leftrightarrow2k\times3k\times5k=-30\)
\(\Leftrightarrow30k^3=-30\)
\(\Leftrightarrow k^3=-1\)
\(\Leftrightarrow k=-1\)
Thay vào ta được :
\(\hept{\begin{cases}x=2k=-2\\y=3k=-3\\z=5k=-5\end{cases}}\)
Vậy ...
\(b,\frac{x}{4}=\frac{y}{3}=\frac{z}{5}\)
\(\Rightarrow\frac{x^2}{16}=\frac{y^2}{9}=\frac{z^2}{25}=\frac{x^2+y^2+z^2}{16+9+25}\)
\(=\frac{200}{50}=4\)
\(\Rightarrow\frac{x^2}{16}=\frac{y^2}{9}=\frac{z^2}{25}=4\)
Đến đây bn tính nốt nhé@_@
a) ta có: \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\Rightarrow\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)
mà xyz = -30 => 2k.3k.5k = - 30
30.k3 = -30
k3 = -1
=> k = - 1
=> x = 2k = 2. (-1) => x = - 2
y = 3k = 3.(-1) => y = - 3
z = 5k = 5.(-1) => z = - 5
KL: x = - 2; y= -3;z = -5
b) ta có: \(\frac{x}{4}=\frac{y}{3}=\frac{z}{5}\Rightarrow\frac{x^2}{16}=\frac{y^2}{9}=\frac{z^2}{25}\)
ADTCDTSBN
có: \(\frac{x^2}{16}=\frac{y^2}{9}=\frac{z^2}{25}=\frac{x^2+y^2+z^2}{16+9+25}=\frac{200}{50}=4\)
\(\Rightarrow\frac{x^2}{16}=4\Rightarrow x^2=64\Rightarrow x=\hept{\begin{cases}8\\-8\end{cases}}\)
\(\frac{y^2}{9}=4\Rightarrow y^2=36\Rightarrow y=\hept{\begin{cases}6\\-6\end{cases}}\)
\(\frac{z^2}{25}=4\Rightarrow z^2=100\Rightarrow z=\hept{\begin{cases}10\\-10\end{cases}}\)
