X=3

Y=8

Z45

Mk chac chan 100%

Đặt \(\frac{x}{8}=\frac{y}{3}=\frac{z}{10}=k\) nên \(\hept{\begin{cases}x=8k\\y=3k\\z=10k\end{cases}}\)

Vì \(xy+yz+zx=1206\) nên \(8k.3k+3k.10k+10k.8k=1206\)

\(\Leftrightarrow24k^2+30k^2+80k^2=1206\)

\(\Leftrightarrow2k^2\left(12+15+40\right)=1206\)

\(\Leftrightarrow2k^2.67=1206\Leftrightarrow2k^2=18\Rightarrow k^2=9\Rightarrow\orbr{\begin{cases}k=3\\k=-3\end{cases}}\)

Với k = 3 thì : \(\frac{x}{8}=\frac{y}{3}=\frac{z}{10}=3\)suy ra x = 24; y = 9 ; z = 30

Với k = - 3 thì \(\frac{x}{8}=\frac{y}{3}=\frac{z}{10}=-3\) suy ra x = - 24; y = - 9 ; z = - 30

Vậy (x;y;z) = { (24;9;30) ; (-24;-9;-30) }

-46/24 5

Ta có:

\(\frac{a}{b}=\frac{c}{d}\)=>\(\frac{3a}{3b}=\frac{3c}{3d}\)=>\(\frac{3a}{3c}=\frac{3b}{3d}\)                                 ;            \(\frac{a}{b}=\frac{c}{d}\)=>\(\frac{4a}{4b}=\frac{4c}{4d}\)=>\(\frac{4a}{4c}=\frac{4b}{4d}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{3a}{3c}=\frac{3b}{3d}=\frac{3a+3b}{3c+3d}\)                                                           ;              \(\frac{4a}{4c}=\frac{4b}{4d}=\frac{4a+4b}{4c+4d}\)

Mà \(\frac{3a}{3b}=\frac{3b}{3d}=\frac{4a}{4c}=\frac{4b}{4d}\)

=>\(\frac{3a+3b}{3c+3d}=\frac{4a+4b}{4c+4d}\)

\(\Rightarrow\)\(\frac{1+2y+1+4y+1+6y}{18+24+6x}\)=\(\frac{\left(1+1+1\right)+2y+4y+6y}{6\left(3+4+x\right)}=\frac{y\left(2+4+6\right)+3}{6\left(3+4+x\right)}=\frac{3+y.12}{6\left(7+x\right)}\)

=\(\frac{3\left(1+4y\right)}{3.2\left(7+x\right)}=\frac{1+4y}{14+2x}\)

\(\Rightarrow\)\(\frac{1}{14}=\frac{2y}{x}\Rightarrow x=14.2y=28y\)

\(\frac{x}{y}=28\)

Giải hộ tôi vs 2 giờ 30 tôi học rồi