Question

a] \(\left\{{}\begin{matrix}2x-y=5\\x^2+xy+y^2=7\end{matrix}\right.\)

b] \(\left\{{}\begin{matrix}x^2-xy+3y^2+2x-5y-4=0\\x+2y=4\end{matrix}\right.\)

Answer 1

b, Ta có : \(\left\{{}\begin{matrix}x^2-xy+3y^2+2x-5y-4=0\\x+2y=4\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x^2-xy+3y^2+2x-5y=4\\x+2y=4\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x^2-xy+3y^2+2x-5y=x+2y\\x+2y=4\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x^2-xy+3y^2+2x-5y-x-2y=0\\x+2y=4\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x^2-xy+3y^2+x-7y=0\\x+2y=4\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x^2+2xy+3y^2+1,5xy-4,5xy+x-7y=0\\x+2y=4\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x\left(x+2y\right)+1,5y\left(x+2y\right)-4,5xy+x-7y=0\\x+2y=4\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}4x+6y-4,5xy+x-7y=0\\x+2y=4\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}5x-y-4,5xy=0\\x+2y=4\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}5\left(4-2y\right)-y-4,5y\left(4-2y\right)=0\\x=4-2y\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}20-10y-y-18y+9y^2=0\\x=4-2y\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}20-29y+9y^2=0\\x=4-2y\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}9y^2-9y-20y+20=0\\x=4-2y\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\left(9y-20\right)\left(y-1\right)=0\\x=4-2y\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}y=1\\y=\frac{20}{9}\end{matrix}\right.\\x=4-2y\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}y=1\\y=\frac{20}{9}\end{matrix}\right.\\\left[{}\begin{matrix}x=4-2.1=4-2=2\\x=4-\frac{2.20}{9}=-\frac{4}{9}\end{matrix}\right.\end{matrix}\right.\)

Vậy phương trình có 2 nghiệm ( x; y ) = \(\left(2;1\right)\), ( x; y ) = \(\left(-\frac{4}{9};\frac{20}{9}\right)\)

Answer 2

a, Ta có : \(\left\{{}\begin{matrix}2x-y=5\\x^2+xy+y^2=7\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=2x-5\\x^2+x\left(2x-5\right)+\left(2x-5\right)^2=7\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=2x-5\\x^2+2x^2-5x+4x^2-20x+25=7\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=2x-5\\7x^2-25x+18=0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=2x-5\\7x^2-7x-18x+18=0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=2x-5\\\left(7x-18\right)\left(x-1\right)=0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=2x-5\\\left[{}\begin{matrix}x=1\\x=\frac{18}{7}\end{matrix}\right.\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}y=2.1-5=2-5=-3\\y=2.\left(\frac{18}{7}\right)-5=\frac{1}{7}\end{matrix}\right.\\\left[{}\begin{matrix}x=1\\x=\frac{18}{7}\end{matrix}\right.\end{matrix}\right.\)

Vậy hệ phương trình trên có 2 nghiệm là ( x; y ) = ( 1; -3 ) , ( x; y ) \(=\left(\frac{18}{7};\frac{1}{7}\right)\)