\(\left\{{}\begin{matrix}xy=80\\-3x+10y=50\end{matrix}\right.< =>\left\{{}\begin{matrix}y=\dfrac{80}{x}\left(1\right)\\-3x+\dfrac{10.80}{x}=50\left(2\right)\end{matrix}\right.\)(x,y\(\ne0\))
giải pt (2) \(-3x+\dfrac{800}{x}=50< =>\dfrac{-3x^2+800}{x}=\dfrac{50x}{x}\)
\(=>-3x^2+800=50x< =>-3x^2-50x+800=0\)
\(\Delta=\left(-50\right)^2-4\left(-3\right)800=12100>0\)
=>\(x1=\dfrac{50+\sqrt{12100}}{2\left(-3\right)}=-\dfrac{80}{3}\)(TM)
\(x2=\dfrac{50-\sqrt{12100}}{2\left(-3\right)}=10\)(TM)
thay \(x1=-\dfrac{80}{3}\)vào pt(1)\(=>y=\dfrac{80}{-\dfrac{80}{3}}=-3\)
thay \(x2=10\) vào pt(1)=>\(y=\dfrac{80}{10}=8\)
vậy hpt có nghiêm (x,y)=\(\left\{\left(-\dfrac{80}{3};-3\right),\left(10,8\right)\right\}\)
