Question

Cho \(x^2+2y^2+2xy+3x+3y-4=0\) Tìm Max, Min của A=x+y

Answer 1

\(x^2+2y^2+2xy+3x+3y-4=0\)

<=> \(x^2+2xy+y^2+3\left(x+y\right)+y^2-4=0\)

<=> \(\left(x+y\right)^2+3\left(x+y\right)-4+y^2=0\)

<=>\(A^2+3A-4+y^2=0\)

<=> (A-1)(A+4)=-y²\(\le0\)

do A-1 <A+4

=> \(\left\{{}\begin{matrix}A-1\le0\\A+4\ge4\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}A\le1\\A\ge-4\end{matrix}\right.\)

<=> \(-4\le A\le1\)

minA xảy ra <=> \(\left\{{}\begin{matrix}y=0\\x+y=-4\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}y=0\\x=-4\end{matrix}\right.\)(t/m)

maxA xảy ra <=> \(\left\{{}\begin{matrix}y=0\\x+y=1\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}y=0\\x=1\end{matrix}\right.\)(t/m)

Answer 2

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