\(A=m^2-2m-5\)

\(=m^2-2m+1-6\)

\(=\left(m-1\right)^2-6\ge-6\)

Dấu '' = '' xảy ra khi \(\left(m-1\right)^2=0\Leftrightarrow m=1\)

Vậy \(Min_A=-6\) khi \(m=1\)

\(A=m^2-2m-5\)

\(=\left(m^2-2m+1\right)-6\)

\(=\left(m-1\right)^2-6\ge-6\left(Vì\left(m-1\right)^2\ge0\forall m\right)\)

Min \(A=-6\Leftrightarrow m=1\)

`A=m^2-2m-5`

`A=m^2-2m+1-6`

`A=(m-1)^2-6`

 Vì `(m-1)^2 >= 0 AA m`

`=>(m-1)^2-6 >= -6 AA m`

 Hay `A >= -6 AA m`

Dấu "`=`" xảy ra `<=>(m-1)^2=0<=>m-1=0<=>m=1`

Vậy `GTN N` của `A` là `-6` khi `m=1`

\(A=\frac{3}{\left(x+2\right)^2+4};\left(x+2\right)^2\in N\)

\(\Rightarrow A_{max}\Leftrightarrow\left(x+2\right)^2=0\Leftrightarrow\left(x+2\right)^2+4=4\)

\(\Rightarrow A_{max}=\frac{3}{4}\)

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

Mặt khác: \(\left(x+1\right)^2;\left(y+3\right)^2\in N\Rightarrow\left(x+1\right)^2+\left(y+3\right)^2\ge0\)

\(\Rightarrow B_{min}\Leftrightarrow\left(x+1\right)^2+\left(y+3\right)^2=0\Rightarrow B_{min}=1\)

\(A=\frac{3}{\left(x+2\right)^2+4}\)

Để A max

=>(x+2)^2+4 min

Mà\(\left(x+2\right)^2\ge0\Rightarrow\left(x+2\right)^2+4\ge4\)

Vậy Min = 4 <=>x=-2

Vậy Max A = 3/4 <=> x=-2

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

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

\(\Rightarrow B\ge0+0+1=1\)

Vậy MinB = 1<=>x=-1;y=-3

a)

\(A=4x-x^2+3=-\left(x^2-4x-3\right)=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)

Daaus = xayr ra khi: x = 2

b) \(B=4x^2-12x+15=4\left(x^2-3x+9\right)-21=4\left(x-3\right)^2-21\ge-21\)

Dấu = xảy ra khi x = 3

c) \(C=4x^2+2y^2-4xy-4y+1=\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3=\left(2x-y\right)^2+\left(y-2\right)^2-3\ge-3\)

Dấu = xảy ra khi

2x = y và y = 2

=> x = 1 và y = 2

a) A = \(-x^2+4x+3=-\left(x-2\right)^2+7\le7\)

Dấu "=" <=> x = 2

b) \(4x^2-12x+15=\left(2x-3\right)^2+6\ge6\)

Dấu "=" xảy ra <=> \(x=\dfrac{3}{2}\)

c) \(4x^2+2y^2-4xy-4y+1\)

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

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

Dấu "=" <=> \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

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 ..........

a. A = 5.(x - 2)² + 1

Ta có: (x - 2)² \(\ge\)0 => 5.(x - 2)² \(\ge\)0 => 5.(x - 2)² + 1 \(\ge\)1

Do đó A có GTNN là 1

<=> x - 2 = 0

<=> x = 2

b. B = 4 - (1/2 - x)²

Ta có: (1/2 - x)² \(\ge\)0

=> 4 - (1/2 - x)² \(\le\)4

Do đó B có GTLN là 4

<=> 1/2 - x = 0

<=> x = 1/2

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