a) Ta có: \(\left(x-2\right)^2\ge0\forall x\)

nên Dấu '=' xảy ra khi x-2=0

hay x=2

Vậy: Gtnn của biểu thức \(\left(x-2\right)^2\) là 0 khi x=2

a, \(A=-x^2-2x+3=-\left(x^2+2x-3\right)=-\left(x^2+2x+1-4\right)\)

\(=-\left(x+1\right)^2+4\le4\)

Dấu ''='' xảy ra khi x = -1 

Vậy GTLN là 4 khi x = -1 

b, \(B=-4x^2+4x-3=-\left(4x^2-4x+3\right)=-\left(4x^2-4x+1+2\right)\)

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

Dấu ''='' xảy ra khi x = 1/2 

Vậy GTLN B là -2 khi x = 1/2 

c, \(C=-x^2+6x-15=-\left(x^2-2x+15\right)=-\left(x^2-2x+1+14\right)\)

\(=-\left(x-1\right)^2-14\le-14\)

Vâỵ GTLN C là -14 khi x = 1

Bài 8 : 

b, \(B=x^2-6x+11=x^2-6x+9+2=\left(x-3\right)^2+2\ge2\)

Dấu ''='' xảy ra khi x = 3

Vậy GTNN B là 2 khi x = 3 

c, \(x^2-x+1=x^2-x+\dfrac{1}{4}+\dfrac{3}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Dấu ''='' xảy ra khi x = 1/2 

Vậy ...

c, \(x^2-12x+2=x^2-12x+36-34=\left(x-6\right)^2-34\ge-34\)

Dấu ''='' xảy ra khi x = 6

Vậy ...

\(A=\frac{3x^2-6x+9}{x^2-2x+3}=3\)

\(A=\left(x^2+2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{5}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\\ A_{min}=-\dfrac{5}{4}\Leftrightarrow x=-\dfrac{3}{2}\\ B=\left(x^2+2xy+y^2\right)+\left(x^2+6x+9\right)+3\\ B=\left(x+y\right)^2+\left(x+3\right)^2+3\ge3\\ B_{min}=3\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-3\end{matrix}\right.\\ C=-\left(x^2-2x+1\right)+1=-\left(x-1\right)^2+1\le1\\ C_{max}=1\Leftrightarrow x=1\)

\(A=x^2-4x+10=x^2-4x+4+6=\left(x-2\right)^2+6\ge6\)

Vậy GTNN A là 6 khi x - 2 = 0 <=> x = 2 

\(B=\left(1-x\right)\left(3x-4\right)=3x-4-3x^2+4x=-3x^2+7x-4\)

\(=-3\left(x^2-\frac{7}{3}x+\frac{4}{3}\right)=-3\left(x^2-2.\frac{7}{6}x+\frac{49}{36}-\frac{1}{36}\right)=-3\left(x-\frac{7}{6}\right)^2+\frac{1}{12}\ge\frac{1}{12}\)

\(=3\left(x-\frac{7}{6}\right)^2-\frac{1}{12}\le-\frac{1}{12}\)Vậy GTLN B là -1/12 khi x = 7/6 

\(C=3x^2-9x+5=3\left(x^2-3x+\frac{5}{3}\right)=3\left(x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{7}{12}\right)\)

\(=3\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\ge-\frac{7}{4}\)Vậy GTNN C là -7/4 khi x = 3/2 

\(D=-2x^2+5x+2=-2\left(x^2-\frac{5}{2}x-1\right)=-2\left(x^2-2.\frac{5}{4}x+\frac{25}{16}-\frac{41}{16}\right)\)

\(=-2\left(x-\frac{5}{4}\right)^2+\frac{21}{8}\le\frac{21}{8}\)Vậy GTLN D là 21/8 khi x = 5/4 

1,\(A=2x^2-6x+7\)

   \(=2\left(x^2-3x+\frac{9}{4}\right)+\frac{5}{2}\)

   \(=2\left(x-\frac{3}{2}\right)^2+\frac{5}{2}\ge\frac{5}{2}\)

Dấu "=" khi \(x=\frac{3}{2}\)

2,\(B=\frac{2x^2-6x+5}{x^2-2x+1}\left(ĐKXĐ:x\ne1\right)\)

\(\Leftrightarrow Bx^2-2Bx+B=2x^2-6x+5\)

\(\Leftrightarrow x^2\left(B-2\right)+2x\left(3-B\right)+B-5=0\)(1) 

*Với B = 2 thì \(\left(1\right)\Leftrightarrow x^2\left(2-2\right)+2x\left(3-2\right)+2-5=0\)

                                \(\Leftrightarrow2x-3=0\)

                                \(\Leftrightarrow x=\frac{3}{2}\left(TmĐKXĐ\right)\)

*Với \(B\ne2\)thì pt (1) là pt bậc 2 ẩn x tham số B

Pt (1) có nghiệm khi \(\Delta\ge0\)

                          \(\Leftrightarrow\left(3-B\right)^2-\left(B-2\right)\left(B-5\right)\ge0\)

                           \(\Leftrightarrow9-6B+B^2-B^2+7B-10\ge0\)

                           \(\Leftrightarrow B\ge1\)

Dấu "=" xảy ra khi \(\left(1\right)\Leftrightarrow-x^2+4x-4=0\)

                                        \(\Leftrightarrow-\left(x-2\right)^2=0\)

                                        \(\Leftrightarrow x=2\left(TmĐKXĐ\right)\)

Thấy 1 < 2 nên B_Min = 1<=> x = 2

Vậy ....

A=(9x²-6x+1)+(7x²+7)-1=(3x²+1)²+7(x²+7)-1

Vì: (3x²+1)²\(\ge\)0 và 7(x²+7)\(\ge\)0

Nên:A\(\ge\) -1

B=\(\frac{A-2}{\left(x-1\right)^2}\)\(\ge\)  -3