a: Ta có: \(A=x^2-2xy+5y^2+4y+51\)

\(=x^2-2xy+y^2+4y^2+4y+1+50\)

\(=\left(x-y\right)^2+\left(2y+1\right)^2+50\ge50\forall x,y\)

Dấu '=' xảy ra khi \(x=y=-\dfrac{1}{2}\)

a) \(A=x^2-2xy+5y^2+4y+51=\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)+50=\left(x-y\right)^2+\left(2y+1\right)^2+50\ge50\)

\(minA=50\Leftrightarrow x=y=-\dfrac{1}{2}\)

c) \(C=\dfrac{9}{-2x^2+4x-7}=\dfrac{9}{-2\left(x^2-2x+1\right)-5}=\dfrac{9}{-2\left(x-1\right)^2-5}\ge\dfrac{9}{-5}=-\dfrac{9}{5}\)

\(minC=-\dfrac{9}{5}\Leftrightarrow x=1\)

d) \(10x^2+4y^2-4xy+8x-4y+20=\left[4y^2-4y\left(x+1\right)+\left(x+1\right)^2\right]+\left(9x^2+6x+1\right)+18=\left(2y-x-1\right)^2+\left(3x+1\right)^2+18\ge18\)

\(minD=18\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{1}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)

e) \(E=9x^2+2y^2+6xy-6x-8y+10=\left[9x^2+6x\left(y-1\right)+\left(y-1\right)^2\right]+\left(y^2-6x+9\right)=\left(3x+y-1\right)^2+\left(y-3\right)^2\ge0\)

\(minE=0\Leftrightarrow\) \(\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=3\end{matrix}\right.\)

a)

Ta có:

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

\(\ge0-2=-2\)

Vậy \(A_{min}=-2\), đạt được khi và chỉ khi \(x-1=0\Leftrightarrow x=1\)

b)\(B=4x^2+4x+8=4x^2+4x+1+7\)

\(=\left(2x+1\right)^2+7\ge0+7=7\)

Vậy \(B_{min}=7\), đạt được khi và chỉ khi \(2x+1=0\Leftrightarrow x=\dfrac{-1}{2}\)

c)

Ta có:

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

\(=2+\dfrac{9}{4}-\left(x^2-2x.\dfrac{3}{2}+\dfrac{9}{4}\right)\)

\(=\dfrac{17}{4}-\left(x-\dfrac{3}{2}\right)^2\le\dfrac{17}{4}-0=\dfrac{17}{4}\)

Vậy \(C_{max}=\dfrac{17}{4}\), đạt được khi và chỉ khi \(x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)

d) Ta có:

\(D=-x^2-5x=-\left(x^2+5x\right)=\dfrac{25}{4}-\left(x^2+2x.\dfrac{5}{2}+\dfrac{25}{4}\right)\)

\(=\dfrac{25}{4}-\left(x+\dfrac{5}{2}\right)^2\le\dfrac{25}{4}-0=\dfrac{25}{4}\)

Vậy \(D_{max}=\dfrac{25}{4}\), đạt được khi và chỉ khi \(x+\dfrac{5}{2}=0\Leftrightarrow x=-\dfrac{5}{2}\)

e) Ta có:

\(E=x^2-4xy+5y^2+10x-22y+28\)

\(=x^2+4y^2+5^2-4xy+10x-20y+y^2-2y+1+2\)

\(=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\)

\(\ge0+0+2=2\)

Vậy \(E_{min}=2\), đạt được khi và chỉ khi \(x-2y+5=y-1=0\)

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

A= -x²+2x+3

=>A= -(x²-2x+3)

=>A= -(x²-2.x.1+1+3-1)

=>A=-[(x-1)²+2]

=>A= -(x+1)²-2

Vì -(x+1)² ≤0=> A≤-2

Dấu "=" xảy ra khi

-(x+1)²=0 => x=-1

Vây A lớn nhất= -2 khi x= -1

B=x²-2x+4y²-4y+8

=> B= (x²-2x+1)+(4y²-4y+1)+6

=> B=(x-1)²+(2y+1)²+6

=> B lớn nhất=6 khi x=1 và y=-1/2

a) \(4x^2+12x+1=\left(4x^2+12x+9\right)-8=\left(2x+3\right)^2-8\ge-8\)

\(ĐTXR\Leftrightarrow x=-\dfrac{3}{2}\)

b) \(4x^2-3x+10=\left(4x^2-3x+\dfrac{9}{16}\right)+\dfrac{151}{16}=\left(2x-\dfrac{3}{4}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\)

\(ĐTXR\Leftrightarrow x=\dfrac{3}{8}\)

c) \(2x^2+5x+10=\left(2x^2+5x+\dfrac{25}{8}\right)+\dfrac{55}{8}=\left(\sqrt{2}x+\dfrac{5\sqrt{2}}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\)

\(ĐTXR\Leftrightarrow x=-\dfrac{5}{4}\)

d) \(x-x^2+2=-\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{9}{4}=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\)

\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)

e) \(2x-2x^2=-2\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{2}=-2\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\le\dfrac{1}{2}\)

\(ĐTXR\Leftrightarrow x=\dfrac{1}{2}\)

f) \(4x^2+2y^2+4xy+4y+5=\left(4x^2+4xy+y^2\right)+\left(y^2+4y+4\right)+1=\left(2x+y\right)^2+\left(y+2\right)^2+1\ge1\)

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

a: Ta có: \(4x^2+12x+1\)

\(=4x^2+12x+9-8\)

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

Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)

b: Ta có: \(4x^2-3x+10\)

\(=4\left(x^2-\dfrac{3}{4}x+\dfrac{5}{2}\right)\)

\(=4\left(x^2-2\cdot x\cdot\dfrac{3}{8}+\dfrac{9}{64}+\dfrac{151}{64}\right)\)

\(=4\left(x-\dfrac{3}{8}\right)^2+\dfrac{151}{16}\ge\dfrac{151}{16}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{8}\)

c: Ta có: \(2x^2+5x+10\)

\(=2\left(x^2+\dfrac{5}{2}x+5\right)\)

\(=2\left(x^2+2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}+\dfrac{55}{16}\right)\)

\(=2\left(x+\dfrac{5}{4}\right)^2+\dfrac{55}{8}\ge\dfrac{55}{8}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{5}{4}\)

\(C=-\left(x^2+4x+4\right)-\left(y^2-8y+16\right)+22\\ =-\left(x^2+2x.2+2^2\right)-\left(y^2-2.y.4+4^2\right)+22\\ =-\left(x+2\right)^2-\left(y-4\right)^2+22\\ Vậy:max_C=22.khi.x=-2.và.y=4\)

Bài 6:

a) Ta có: \(A=-x^2+6x-11\)

\(=-\left(x^2-6x+11\right)\)

\(=-\left(x-3\right)^2-2\le-2\forall x\)

Dấu '=' xảy ra khi x=3

b) Ta có: \(B=-x^2-8x+5\)

\(=-\left(x^2+8x-5\right)\)

\(=-\left(x^2+8x+16-21\right)\)

\(=-\left(x+4\right)^2+21\le21\forall x\)

Dấu '=' xảy ra khi x=-4

c) Ta có: \(C=-x^2+4x+1\)

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

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

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

Dấu '=' xảy ra khi x=2

Bài 7:

a) Ta có: \(x^2-6x+11\)

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

\(=\left(x-3\right)^2+2\ge2\forall x\)

Dấu '=' xảy ra khi x=3