\(a,-x^2+2x+5=-\left(x^2-2x-5\right)=-\left(x^2-2x+1-6\right)=-\left(x-1\right)^2+6\le6\)

dấu'=' xảy ra<=>x=1=>Max A=6

\(b,B=-x^2-y^2+4x+4y+2=-x^2+4x-4-y^2+4x-4+10\)

\(=-\left(x^2-4x+4\right)-\left(y^2-4x+4\right)+10\)

\(=-\left(x-2\right)^2-\left(y-2\right)^2+10=-\left[\left(x-2\right)^2+\left(y-2\right)^2\right]+10\le10\)

dấu"=" xảy ra<=>x=y=2=>Max B=10

\(c,C=x^2+y^2-2x+6y+12=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\)

dấu'=' xảy ra<=>x=1,y=-3=>MinC=2

b) Ta có: \(B=x^2+2x+y^2-4y+6\)

\(=x^2+2x+1+y^2-4y+4+1\)

\(=\left(x+1\right)^2+\left(y-2\right)^2+1\ge1\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

Vậy: \(B_{min}=1\) khi (x,y)=(-1;2)

c) Ta có: \(C=4x^2+4x+9y^2-6y-5\)

\(=4x^2+4x+1+9y^2-6y+1-7\)

\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(C_{min}=-7\) khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)

\(A=2x^2+x=2\left(x^2+\dfrac{1}{2}x\right)=2\left(x^2+2.\dfrac{1}{4}x+\dfrac{1}{16}-\dfrac{1}{16}\right)\)

\(=2\left[\left(x+\dfrac{1}{4}\right)^2-\dfrac{1}{16}\right]\ge-\dfrac{1}{8}\) dấu"=' xảy ra<=>x=\(-\dfrac{1}{4}\)

\(B=x^2+2x+y^2-4y+6\)

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

\(\ge1\) dấu"=" xảy ra<=>x=-1;y=2

\(C=4x^2+4x+9y^2-6y-5\)

\(=4x^2+4x+1+9y^2-6y+1-7\)

\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\)

dấu"=" xảy ra<=>x=\(-\dfrac{1}{2},y=\dfrac{1}{3}\)

\(D=\left(2+x\right)\left(x+4\right)-\left(x-1\right)\left(x+3\right)^2\)

=\(x^2+6x+8-\left(x-1\right)\left(x+3\right)^2\)

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

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

dấu"=" xảy ra\(< =>\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

a, xem lại đề 

\(b,x^2-4x+y^2-6y+1\\ =\left(x^2-4x+4\right)+\left(y^2-6y+9\right)-12\\ =\left(x-2\right)^2+\left(y-3\right)^2-12\ge-12\)

Dấu "=" xảy ra\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)

Vậy ...

\(c,x^2-4xy+5y^2-2y+5\\ =\left(x^2-4xy+4y^2\right)+\left(y^2-2y+1\right)+4\\ =\left(x-2y\right)^2+\left(y-1\right)^2+4\ge4\)

Dấu "=" xảy ra\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy ...

a, 

Dấu "=" xảy ra⇔{x=2y=3⇔{x=2y=3

Vậy ...

Dấu "=" xảy ra⇔{x=2y=1⇔{x=2y=1

Vậy ...

viết lại đề đi

\(A=25x^2-20x+7\)

\(\Leftrightarrow A=\left(5x-2\right)^2+3\ge3\)

Dấu " = " xảy ra \(\Leftrightarrow5x-2=0\Leftrightarrow x=\frac{2}{5}\)

Vậy \(minA=3\Leftrightarrow x=\frac{2}{5}\)

\(B=-x^2+2x-2\)

\(\Leftrightarrow B=-\left(x^2-2x+1\right)-3\)

\(\Leftrightarrow B=-\left(x-1\right)^2-3\le-3\)

Dấu " = " xảy ra \(\Leftrightarrow x=1\)

Vậy \(maxB=-3\Leftrightarrow x=1\)

\(C=9x^2-12x\)

\(\Leftrightarrow C=\left(9x^2-12x+4\right)-4\)

\(\Leftrightarrow C=\left(3x-2\right)^2-4\ge-4\)

Dấu " = " xảy ra \(\Leftrightarrow3x-2=0\Leftrightarrow x=\frac{2}{3}\)

Vậy \(minC=-4\Leftrightarrow x=\frac{2}{3}\)

\(D=3-10x^2-4xy-4y^2\)

\(\Leftrightarrow D=-\left(4y^2+4xy+x^2+9x^2\right)-3\)

\(\Leftrightarrow D=-\left[\left(2y-x\right)^2+3x^2\right]-3\le-3\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}2y-x=0\\3x^2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y=0\\x=0\end{cases}}\)

Vậy \(maxD=-3\Leftrightarrow x=y=0\)

\(E=4x-x^2+1\)

\(\Leftrightarrow E=-\left(x^2-4x+4\right)+5\)

\(\Leftrightarrow E=-\left(x-2\right)^2+5\le5\)

Dấu " = " xảy ra \(\Leftrightarrow x=2\)

Vậy \(maxE=5\Leftrightarrow x=2\)

a) Ta có: \(25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3\ge3\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{2}{5}\)

b) Ta có: \(9x^2-6x+2\)

\(=9x^2-6x+1+1\)

\(=\left(3x-1\right)^2+1\ge1\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{3}\)

c) Ta có: \(-x^2+2x-2\)

\(=-\left(x^2-2x+2\right)\)

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

\(=-\left(x-1\right)^2-1\le-1\forall x\)

Dấu '=' xảy ra khi x-1=0

hay x=1

d) Ta có: \(x^2+12x+39\)

\(=x^2+12x+36+3\)

\(=\left(x+6\right)^2+3\ge3\forall x\)

Dấu '=' xảy ra khi x=-6

e) Ta có: \(-x^2-12x\)

\(=-\left(x^2+12x+36-36\right)\)

\(=-\left(x+6\right)^2+36\le36\forall x\)

Dấu '=' xảy ra khi x=-6

f) Ta có: \(4x-x^2+1\)

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

\(=-\left(x^2-4x+4-5\right)\)

\(=-\left(x-2\right)^2+5\le5\forall x\)

Dấu '=' xảy ra khi x=2

a) Ta có: \(25x^2-20x+7\)

\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)

\(=\left(5x-2\right)^2+3\ge3\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{2}{5}\)

b) Ta có: \(9x^2-6x+2\)

\(=9x^2-6x+1+1\)

\(=\left(3x-1\right)^2+1\ge1\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{3}\)

c) Ta có: \(-x^2+2x-2\)

\(=-\left(x^2-2x+2\right)\)

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

\(=-\left(x-1\right)^2-1\le-1\forall x\)

Dấu '=' xảy ra khi x=1

( Mình trình bày mẫu câu a các câu khác mình làm tắt lại nhưng tương tự trình bày câu a nha )

a, Ta có : \(25x^2-20x+7=\left(5x\right)^2-2.5x.2+2^2+3\)

\(=\left(5x-2\right)^2+3\)

Thấy : \(\left(5x-2\right)^2\ge0\forall x\in R\)

\(\Rightarrow\left(5x-2\right)^2+3\ge3\forall x\in R\)

Vậy \(Min=3\Leftrightarrow5x-2=0\Leftrightarrow x=\dfrac{2}{5}\)

b, \(=9x^2-2.3x+1+1=\left(3x-1\right)^2+1\ge1\)

Vậy Min = 1 <=> x = 1/3

c, \(=-x^2+2x-1-1=-\left(x^2-2x+1\right)-1=-\left(x-1\right)^2-1\le-1\)

Vậy Max = -1 <=> x = 1

d, \(=x^2+2.x.6+36+3=\left(x+6\right)^2+3\ge3\)

Vậy Min = 3 <=> x = - 6

e, \(=-x^2-2.x.6-36+36=-\left(x+6\right)^2+36\le36\)

Vậy Max = 36 <=> x = -6 .

f, \(=-x^2+4x-4+5=-\left(x^2-4x+4\right)+5=-\left(x-2\right)^2+5\le5\)

Vậy Max = 5 <=> x = 2

\(a,M=x^2-4x+5=\left(x-2\right)^2+5\\ \Rightarrow M\ge5\)

Dấu "=" xảy ra \(\Leftrightarrow x=2\)

\(b,N=y^2-y-3=\left(y-\dfrac{1}{2}\right)^2-\dfrac{13}{4}\\ \Rightarrow N\ge-\dfrac{13}{4} \)

Dấu "=" xảy ra \(\Leftrightarrow y=\dfrac{1}{2}\)

\(P=x^2+y^2-4x+y+7=\left(x-2\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\\ \Rightarrow P\ge\dfrac{11}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\dfrac{1}{2}\end{matrix}\right.\)

a: M=x^2-4x+4+1

=(x-2)^2+1>=1

Dấu = xảy ra khi x=2

b: N=y^2-y+1/4-13/4

=(y-1/2)^2-13/4>=-13/4

Dấu = xảy ra khi y=1/2

c: P=x^2-4x+4+y^2+y+1/4+11/4

=(x-2)^2+(y+1/2)^2+11/4>=11/4

Dấu = xảy ra khi x=2 và y=-1/2

a) \(=\left(x-2\right)^2\)

b) \(=\left(3x-2\right)^2\)

c) \(=\left(x-3y\right)^2\)

d) \(=\left(\dfrac{x}{2}+1\right)^2\)

e) \(=\left(x-4\right)^2\)

f) \(=\left(\dfrac{1}{2}xy^2+1\right)^2\)

g) \(=\left(x-1\right)\left(x+1\right)\)

h) \(=\left(5x-4\right)\left(5x+4\right)\)

a) x²-4x+4=\(\left(X-2\right)^2\)

b)9x²-12x+4=\(\left(3x-2\right)^2\)

c)x²-6xy+9y^{2=\(\left(x-3y\right)^2\)}

d)x²/4+x+1=\(\left(\dfrac{x}{2}+1\right)^2\)

e)-8x+16+x² =\(X^2-8x+16=\left(x-4\right)^2\)

f)xy²+1/4x²y⁴+1 => sai đề

g)x²-1=\(\left(X-1\right)\left(x+1\right)\)

h)25x²-16=\(\left(5x-4\right)\left(5x+4\right)\)