\(\left\{{}\begin{matrix}\dfrac{72}{x}+\dfrac{54}{y}=6\\x-y=6\end{matrix}\right.\left(x,y>0\right)< =>\left\{{}\begin{matrix}x=6+y\left(1\right)\\\dfrac{72}{\left(6+y\right)}+\dfrac{54}{y}=6\left(2\right)\end{matrix}\right.\)(x,y>0,y\(\ne-6\))
giải pt(2) \(\dfrac{72}{\left(6+y\right)}+\dfrac{54}{y}=6< =>\dfrac{72y+54\left(6+y\right)}{y\left(6+y\right)}=6\)
\(< =>\dfrac{126y+324}{y\left(6+y\right)}=6=>126y+324=6y\left(6+y\right)\)
\(< =>126y+324=36y+6y^2\)
\(< =>-6y^2+90y+324=0\)
\(\Delta=90^2-4\left(-6\right).324=15876>0\)
=>x1=\(\dfrac{-90+\sqrt{15876}}{2\left(-6\right)}=-3\left(loai\right)\)
x2=\(\dfrac{-90-\sqrt{15876}}{2\left(-6\right)}=18\left(TM\right)\)
=>x=x2=18 thay vào pt(1)=>x=6+18=24
vậy (x,y)=(24,18)
