a)

\(A=4x-x^2+3=-\left(x^2-4x-3\right)=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

Daaus = xayr ra khi: x = 2

b) \(B=4x^2-12x+15=4\left(x^2-3x+9\right)-21=4\left(x-3\right)^2-21\ge-21\)

Dấu = xảy ra khi x = 3

c) \(C=4x^2+2y^2-4xy-4y+1=\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3=\left(2x-y\right)^2+\left(y-2\right)^2-3\ge-3\)

Dấu = xảy ra khi

2x = y và y = 2

=> x = 1 và y = 2

a) A = \(-x^2+4x+3=-\left(x-2\right)^2+7\le7\)

Dấu "=" <=> x = 2

b) \(4x^2-12x+15=\left(2x-3\right)^2+6\ge6\)

Dấu "=" xảy ra <=> \(x=\dfrac{3}{2}\)

c) \(4x^2+2y^2-4xy-4y+1\)

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

= \(\left(2x-y\right)^2+\left(y-2\right)^2-3\ge-3\)

Dấu "=" <=> \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

a) \(C=4x^2+3y^2+4xy-4x-10y+7=\left[4x^2+4x\left(y-1\right)+\left(y-1\right)^2\right]+2\left(y^2-4y+4\right)-2=\left(2x+y-1\right)^2+2\left(y-2\right)^2-2\ge-2\)

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

d) \(D=x^2-2xy+6y^2-12x+2y+45=\left[x^2-2x\left(y+6\right)+\left(y+6\right)^2\right]+5\left(y^2-2y+1\right)+4=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\ge4\)

\(minD=4\Leftrightarrow\) \(\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)

Ta có:

A = 12x - 4x² + 3 = -4(x² - 3x + 9/4) + 12 = -4(x - 3/2)² + 12

Ta luôn có: (x - 3/2)² \(\ge\)0 \(\forall\)x => -4(x  - 3/2)² \(\le\)0 \(\forall\)x

=> -4(x - 3/2)² + 12 \(\le\)12 \(\forall\)x

Dấu "=" xảy ra khi : (x - 3/2)² = 0 <=> x - 3/2 = 0 <=> x = 3/2

Vậy A_max = 12 tại x= 3/2

\(C=6x-x^2+3\)

\(C=-\left(x^2-6x+9\right)+12\)

\(C=-\left(x-3\right)^2+12\)

\(\le12\)

Dấu "=" xảy ra khi \(x=3\)

\(D=2x-6y-x^2-y^2-2\)

\(D=-\left(x^2-2x+1\right)-\left(y^2-6y+9\right)+6\)

\(D=-\left(x-1\right)^2-\left(y-3\right)^2+6\)

\(\le6\)

Dấu "=" xảy ra khi \(x=1;y=3\)

a) A = x² - 6x + 13 = x² - 2.x.3 + 3³ +4 = (x-3)^{2 +} 4 >= 4 suy ra minA=4 
mấy câu kia giải tương tự

a) \(A=-9x^2+12x+2\)

\(A=-9x^2+12x-4+6\)

\(A=6-\left(3x-2\right)^2\)

Có: \(\left(3x-2\right)^2\ge0\Rightarrow6-\left(3x-2\right)^2\le6\)

Dấu = xảy ra khi: \(\left(3x-2\right)^2=0\Rightarrow3x-2=0\Rightarrow x=\frac{2}{3}\)

Vậy: \(Max_A=6\) tại \(x=\frac{2}{3}\)

\(A=\left(x^2-4x+4\right)+4=\left(x-2\right)^2+4\ge4\)

\(minA=4\Leftrightarrow x=2\)

\(B=\left(4x^2-12x+9\right)+2=\left(2x-3\right)^2+2\ge2\)

\(minB=2\Leftrightarrow x=\dfrac{3}{2}\)

\(C=3\left(x^2+2x+1\right)-8=3\left(x+1\right)^2-8\ge-8\)

\(minC=-8\Leftrightarrow x=-1\)

\(D=-\left(x^2-2x+1\right)-4=-\left(x-1\right)^2-4\le-4\)

\(maxD=-4\Leftrightarrow x=1\)

\(E=-\left(4x^2-6x+\dfrac{9}{4}\right)-\dfrac{11}{4}=-\left(2x-\dfrac{3}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\)

\(maxA=-\dfrac{11}{4}\Leftrightarrow x=\dfrac{3}{4}\)

\(F=-2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{55}{8}=-2\left(x-\dfrac{1}{4}\right)^2-\dfrac{55}{8}\le-\dfrac{55}{8}\)

\(maxF=-\dfrac{55}{8}\Leftrightarrow x=\dfrac{1}{4}\)

\(G=\left(x^2-4xy+4y^2\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-2y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

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

\(H=-\left(x^2-2x+1\right)-\left(y^2+4y+4\right)+16=-\left(x-1\right)^2-\left(y+2\right)^2+16\le16\)

\(maxH=16\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

C= ( 2x+y)^2 + 2(2x+y) + 1 + y^2 + 4y +4 + 12

C= (2x+y+1)^2 +( y+2)^2 + 12

Từ đó suy ra min C là 12 khi y = -2; x= 1/2