a) \(x^2-3x=4\)
\(\Leftrightarrow x^2-3x-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\) \(\left(a-b+c=0\right)\)
b) \(\left\{{}\begin{matrix}2x-5-y=0\\5x+3y=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=5\\5x+3y=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-3y=15\\5x+3y=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}11x=33\\y=2x-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(3;1\right)\)
`a)x^2 -3x=4`
`<=>x^2 -3x-4=0`
`<=> x^2 -4x+x-4=0`
`<=> (x-4)(x+1)=0`
`<=>` \(\left[{}\begin{matrix}x-4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-1\end{matrix}\right.\)
Vậy `S={4;-1}`
`b)`
\(\left\{{}\begin{matrix}2x-5-y=0\\5x+3y=18\end{matrix}\right.\)
`<=>` \(\left\{{}\begin{matrix}2x-y=5\\5x+3y=18\end{matrix}\right.\)
`<=>` \(\left\{{}\begin{matrix}6x-3y=15\\5x+3y=18\end{matrix}\right.\)
`<=>` \(\left\{{}\begin{matrix}11x=33\\2x-y=5\end{matrix}\right.\)
`<=>` \(\left\{{}\begin{matrix}x=3\\2.3-y=5\end{matrix}\right.\)
`<=>` \(\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)
Vậy `(x;y)=(3;1)`
