e: Đặt \(\frac{x}{8}=\frac{y}{3}=\frac{z}{10}=k\)
=>x=8k; y=3k; z=10k
xy+yz+xz=1206
=>\(8k\cdot3k+8k\cdot10k+3k\cdot10k=1206\)
=>\(24k^2+80k^2+30k^2=1206\)
=>\(134k^2=1206\)
=>\(k^2=9\)
=>k=3 hoặc k=-3
TH1: k=3
=>\(x=8\cdot3=24;y=3\cdot3=9;z=10\cdot3=30\)
TH2: k=-3
=>\(x=8\cdot\left(-3\right)=-24;y=3\cdot\left(-3\right)=-9;z=10\cdot\left(-3\right)=-30\)
f: Ta có: \(\frac{x}{4}=\frac{2y}{5}=\frac{5z}{6}\)
=>\(60\cdot\frac{x}{4}=60\cdot\frac{2y}{5}=60\cdot\frac{5z}{6}\)
=>15x=24y=50z
=>\(\frac{15x}{600}=\frac{24y}{600}=\frac{50z}{600}\)
=>\(\frac{x}{40}=\frac{y}{25}=\frac{z}{12}\)
Đặt \(\frac{x}{40}=\frac{y}{25}=\frac{z}{12}=k\)
=>x=40k; y=25k; z=12k
\(x^2-3y^2+2z^2=325\)
=>\(\left(40k\right)^2-3\cdot\left(25k\right)^2+2\cdot\left(12k\right)^2=325\)
=>\(1600k^2-1875k^2+288k^2=325\)
=>\(13k^2=325\)
=>\(k^2=25\)
=>k=5 hoặc k=-5
TH1: k=5
=>\(\begin{cases}x=40\cdot5=200\\ y=25\cdot5=125\\ z=12\cdot5=60\end{cases}\)
TH2: k=-5
=>\(\begin{cases}x=40\cdot\left(-5\right)=-200\\ y=25\cdot\left(-5\right)=-125\\ z=12\cdot\left(-5\right)=-60\end{cases}\)
g: Đặt \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=k\)
=>x-1=2k; y-2=3k; z-3=4k
=>x=2k+1; y=3k+2; z=4k+3
x-2y+3z=14
=>2k+1-2(3k+2)+3(4k+3)=14
=>2k+1-6k-4+12k+9=14
=>8k+6=14
=>8k=8
=>k=1
=>\(\begin{cases}x=2\cdot1+1=2+1=3\\ y=3\cdot1+2=5\\ z=4\cdot1+3=7\end{cases}\)
h: 2x=3y
=>\(\frac{x}{3}=\frac{y}{2}\)
=>\(\frac{x}{21}=\frac{y}{14}\) (1)
5y=7z
=>\(\frac{y}{7}=\frac{z}{5}\)
=>\(\frac{y}{14}=\frac{z}{10}\left(2\right)\)
Từ (1),(2) suy ra \(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)
mà 3x-7y+5z=30
nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{3x-7y+5z}{3\cdot21-7\cdot14+5\cdot10}=\frac{30}{15}=2\)
=>\(\begin{cases}x=2\cdot21=42\\ y=2\cdot14=28\\ z=2\cdot10=20\end{cases}\)
