Giải:
Ta có:
\(yz.zt=24.32\)
\(yt.z^2=24.32\)
\(48.z^2=24.32\)
\(\Rightarrow z^2=\dfrac{24.32}{48}=16\)
\(\Rightarrow z=4\)
Ta có:
\(yz=24\)
\(y.4=24\)
\(\Rightarrow y=6\)
Ta có:
\(xy=12\)
\(x.6=12\)
\(\Rightarrow x=2\)
Ta có:
\(y.t=48\)
\(6.t=48\)
\(\Rightarrow t=48:6=8\)
Vậy:
\(x=2\) , \(y=6\) , \(z=4\) , \(t=8\) .
\(\left\{{}\begin{matrix}yt=48\\yz=24\\xy=12\\zt=32\end{matrix}\right.\)
Nhân hết lại: \(\left(yt\right)\left(yz\right)\left(xy\right).\left(zt\right)=48.24.12.32\)
Ghép lại VP: \(\left(zt\right)^2.\left(xy\right).y^2=48.24.12.32\)
Vậy thừa ra y^2: \(y^2=\dfrac{48.24.12.32}{32^2.12}=\dfrac{24.48}{32}=\dfrac{8.3.4.12}{8.4}=36\)
\(\Rightarrow\left[{}\begin{matrix}y=-6\\y=6\end{matrix}\right.\)
Thay vào từng cái trên có:
\(\left\{{}\begin{matrix}y=6\\t=8\\z=4\\x=2\end{matrix}\right.\) \(\left\{{}\begin{matrix}y=-6\\t=-8\\z=-4\\x=-2\end{matrix}\right.\)
Kết luận: (x,y,z,t)=(2,6,4,8) ;(-2,-6,-4,-8)
