Question

[Hà NgọcThanhHY]

Bài 31 vs 32 ạ

Answer 1

[Lấp La Lấp Lánh]

31) 

a) \(A=x^2+y^2-2x+4y+15=\left(x^2-2x+1\right)+\left(y^2+4x+4\right)+10=\left(x-1\right)^2+\left(y+2\right)^2+10\ge10\)

\(minA=10\Leftrightarrow\)\(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

b) \(B=x^2+y^2-x+6y+20=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+6y+9\right)+\dfrac{43}{4}=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{43}{4}\ge\dfrac{43}{4}\)

\(minB=\dfrac{43}{4}\Leftrightarrow\)\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-3\end{matrix}\right.\)

c) \(C=2x^2+5y^2+4xy+8x-4y+15=\left(x^2+4xy+4y^2\right)+\left(x^2+8x+16\right)+\left(y^2-4y+4\right)-5=\left(x+2y\right)^2+\left(x+4\right)^2+\left(y-2\right)^2-5\ge-5\)

\(minC=-5\Leftrightarrow\)\(\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\)

32)

a) \(A=-x^2+6x+27=-\left(x^2-6x+9\right)+36=-\left(x-3\right)^2+36\le36\)

\(maxA=36\Leftrightarrow x=3\)

b) \(B=-9x^2-6x+19=-\left(9x^2+6x+1\right)+20=-\left(3x+1\right)^2+20\le20\)

\(maxB=20\Leftrightarrow x=-\dfrac{1}{3}\)

c) \(C=12x-4x^2+3=-\left(4x^2-12x+9\right)+12=-\left(2x-3\right)^2+12\le12\)

\(maxC=12\Leftrightarrow x=\dfrac{3}{2}\)