\(M=\frac{x^2+1}{x^2-x+1}\)\(\Rightarrow M\left(x^2-x+1\right)=x^2+1\)

\(\Leftrightarrow Mx^2-Mx+M-x^2-1=0\)

\(\Leftrightarrow x^2\left(M-1\right)-Mx-M+1=0\)

\(\Delta=\left(-M\right)^2-4\left(M-1\right)^2\ge0\)

\(\Leftrightarrow M^2-4\left(M^2-2M+1\right)\ge0\)

\(\Leftrightarrow-3M^2+8M-4\ge0\)

\(\Leftrightarrow3M^2-8M+4\le0\)

\(\Leftrightarrow\left(3M-2\right)\left(M-2\right)\le0\)

\(\Leftrightarrow\frac{2}{3}\le M\le2\)

Vậy \(Min_M=\frac{2}{3}\Leftrightarrow x=\frac{-b}{2a}=\frac{M}{2\left(M-1\right)}=\frac{\frac{2}{3}}{2\left(\frac{2}{3}-1\right)}=-1\)

\(Max_M=2\Leftrightarrow x=\frac{-b}{2a}=\frac{M}{2\left(M-1\right)}=\frac{2}{2\left(2-1\right)}=1\)

GTNN của A:

A=x²+1/x²-x+1=1+x/x²+1-x

=>A>1

suy ra:GTNN cùa A=2 với x=1

A=2

X=1

ta có |x+19|+|y-5|+1980 >1980

<=>|x+19|+|y-5|>0

dấu"="chỉ xảy ra <=>|x+19|=0vs|y-5|=0<=>x+19=0vsy-5=0

                                   <=>x=-19,y=5

1 )Vì \(\left(x+2\right)^2\ge0;\left(y-3\right)^2\ge0\)

\(\Rightarrow\left(x+2\right)^2+\left(y-3\right)^2\ge0\)

\(\Rightarrow\left(x+2\right)^2+\left(y-3\right)^2+1\ge1\)

Dấu "=: xảy ra <=> \(\orbr{\begin{cases}\left(x+2\right)^2=0\\\left(y-3\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\y=3\end{cases}}}\)

Vậy ........

2 ) \(\frac{1}{\left(x-2\right)^2+2}\ge\frac{1}{2}\)

Dấu "=" xảy ra <=> x = 2

Vậy ..........

+) Min: \(A=\frac{x^2}{x^4+x^2+1}\ge0\forall x\) 

Dấu "=" <=> x=0

+) Max: \(1-3A=\frac{x^4-2x^2+1}{x^4+x^2+1}=\frac{\left(x^2-1\right)^2}{x^4+x^2+1}\ge0\)

\(\Rightarrow A\le\frac{1}{3}\)Dấu "=" <=> x= 1,-1

GTNN 

Xét tử : x^4+x^2+5= x^4+2x^2+1 -x^2+4 =(x^2+1)^2 -(x-2)(x+2)

=> GTNN của Biểu thức là 1 <=> x=2 hoặc x= -2

GTLN: Ko có