\(\dfrac{3x}{8}=\dfrac{y}{4}=\dfrac{3z}{16}\)
=>\(\dfrac{3x}{8}\cdot8=\dfrac{y}{4}\cdot8=\dfrac{3z}{16}\cdot8\)
=>\(3x=2y=1,5z\)
=>\(\dfrac{3x}{6}=\dfrac{2y}{6}=\dfrac{1.5z}{6}\)
=>\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=k\)
=>x=2k; y=3k; z=4k
\(2x^2+2y^2-z^2=10\)
=>\(2\cdot\left(2k\right)^2+2\cdot\left(3k\right)^2-\left(4k\right)^2=10\)
=>\(8k^2+18k^2-16k^2=10\)
=>\(10k^2=10\)
=>\(k^2=1\)
=>\(\left[{}\begin{matrix}k=1\\k=-1\end{matrix}\right.\)
TH1: k=1
=>\(x=2\cdot1=2;y=3\cdot1=3;z=4\cdot1=4\)
TH2: k=-1
=>\(x=2\cdot\left(-1\right)=-2;y=3\cdot\left(-1\right)=-3;z=4\cdot\left(-1\right)=-4\)
Đặt `(3x)/8 = y/4 = (3z)/16 = k`
`=> 3x = 8k; y = 4k; 3z = 16k`
`=> x = 8/3 k; y = 4k; z = 16/3 k`
Khi đó: `2x^2 + 2y^2 - z^2 = 10`
`<=> 2(8/3 k)^2 + 2 (4k)^2 - (16/3 k )^2 = 10`
`<=> 128/9 k^2 + 32 k^2 - 256/9 k^2 = 10`
`<=> (128/9 + 32 - 256/9) k^2 = 10`
`<=> 160/9 k^2 = 10`
`<=> k^2 = 9/16`
`<=> k^2 = (3/4)^2 `
`<=> k = 3/4` hoặc `k = -3/4`
Xét ` k = 3/4` thì:
`x = 8/3 . 3/4 = 2`
`y = 4 . 3/4 = 3`
`z = 16/3 . 3/4 = 4`
Xét `k = -3/4` thì:
`x = 8/3 . (-3/4) = -2`
`y = 4 . (-3/4) = -3`
`z = 16/3 . (03/4) = -4`
Vây ...