\(a.A=\left(x-2\right)^2+\left(y+1\right)^2+1\ge1\forall x;y\) . " = " \(\Leftrightarrow x=2;y=-1\) 

b.\(B=7-\left(x+3\right)^2\le7\forall x\)  " = " \(\Leftrightarrow x=-3\)

c.\(C=\left|2x-3\right|-13\ge-13\forall x\)  " = " \(\Leftrightarrow x=\dfrac{3}{2}\)

d.\(D=11-\left|2x-13\right|\le11\forall x\)  " = " \(\Leftrightarrow x=\dfrac{13}{2}\)

\(A=\left(x-1\right)^2+8\ge8\\ A_{min}=8\Leftrightarrow x=1\\ B=\left(x+3\right)^2-12\ge-12\\ B_{min}=-12\Leftrightarrow x=-3\\ C=x^2-4x+3+9=\left(x-2\right)^2+8\ge8\\ C_{min}=8\Leftrightarrow x=2\\ E=-\left(x+2\right)^2+11\le11\\ E_{max}=11\Leftrightarrow x=-2\\ F=9-4x^2\le9\\ F_{max}=9\Leftrightarrow x=0\)

giải nhanh đi nhé mik cần gấp ai lm đủ đúng hết mik k mun cho nha giải đủ các bước nhé cảm ưn các bạn trước giúp mik nha^.^><hihiii

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

vi \(\left(x+1\right)^2\ge0\)(voi moi x)

    \(\Rightarrow\left(x+1\right)^2+2\ge2\)(voi moi x)

Vay GTNN cua A =2 khi x=-1

2)  Goi 2 so nguyen lien tiep do la x va x+1

TDTC x+1-x=1

Vi 1 la so le nen x+1-x la so le 

Vay .......

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

\(=-2y\cdot2x=-4xy\)(dpcm)

4) \(Q=-x^2+6x+1=-\left(x^2-6x-1\right)=-\left(x^2-6x+9-10\right)=-\left(x-3\right)^2+10\)

Vi \(\left(x-3\right)^2\ge0\)(voi moi x)

\(\Rightarrow-\left(x-3\right)^2\le0\)(voi moi x)

\(\Rightarrow-\left(x-3\right)^2+10\le10\)(voi moi x)

Vay GTLN cua Q=10 khi x=3

a)-19

b)22

\(A=3\left(x-3\right)^2+\left(y-1\right)^2+2005\)

Nhận xét: \(\left(x-3\right)^2\ge0\forall x\)\(\Rightarrow3\left(x-3\right)^2\ge0\forall x\)

                \(\left(y-1\right)^2\ge0\forall y\)

\(\Rightarrow3\left(x-3\right)^2+\left(y-1\right)^2\ge0\forall x,y\)

\(\Rightarrow3\left(x-3\right)^2+\left(y-1\right)^2+2005\ge2005\forall x,y\)

Vậy \(minA=2005\)khi   \(3\left(x-3\right)^2=0\)\(\Rightarrow x-3=0\)\(\Rightarrow x=3\)

                                                 \(\left(y-1\right)^2=0\)\(\Rightarrow y-1=0\)\(\Rightarrow y=1\)

KL: Vậy \(minA=2005\) khi  \(x=3;y=1\)

\(B=\left(x^2-9\right)^2+|y-2|-1\)

Nhận xét:  \(\left(x^2-9\right)^2\ge0\forall x\)

                  \(|y-2|\ge0\forall y\)

\(\Rightarrow\left(x^2-9\right)^2+|y-2|\ge0\forall x,y\)

\(\Rightarrow\left(x^2-9\right)^2+|y-2|-1\ge-1\forall x,y\)

Vậy \(minB=-1\)khi   \(\left(x^2-9\right)^2=0\)\(\Rightarrow x^2-9=0\)\(\Rightarrow x^2=9\)\(\Rightarrow x=3\)

                                              \(|y-2|=0\)\(\Rightarrow y=2\)

KL: Vậy \(minB=-1\) khi  \(x=3;y=2\)

\(C=x^2-2x+5\)

\(\Rightarrow C=x^2-2x+1+4\)

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

Nhận xét: \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-1\right)^2+4\ge4\forall x\)

Vậy  \(minB=4\) khi  \(\left(x-1\right)^2=0\)\(\Rightarrow x-1=0\)\(\Rightarrow x=1\)

KL: Vậy \(minB=4\) khi  \(x=1\)