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

Bài 1:

a: \(M=x^2-10x+3\)

\(=x^2-10x+25-22\)

\(=\left(x^2-10x+25\right)-22\)

\(=\left(x-5\right)^2-22>=-22\forall x\)

Dấu '=' xảy ra khi x-5=0

=>x=5

b: \(N=x^2-x+2\)

\(=x^2-x+\dfrac{1}{4}+\dfrac{7}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\forall x\)

Dấu '=' xảy ra khi x-1/2=0

=>x=1/2

c: \(P=3x^2-12x\)

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

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

\(=3\left(x-2\right)^2-12>=-12\forall x\)

Dấu '=' xảy ra khi x-2=0

=>x=2

\(A=x^2-6x+15=\left(x^2-6x+9\right)+6\)

\(=\left(x-3\right)^2+6\ge6\)

\(minA=6\Leftrightarrow x=3\)

A=x²-2x3+3²+6

A=(x-3)²+6

Vì (x-3)² luôn > hoặc = 0 với mọi x

=> (x-3)²+6 > hoặc = 6

Vậy GTNN = 6 

Dấu "=" xảy ra khi x-3=0

X=3

\(A=x^2-6x+15\)

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

\(\Rightarrow A=\left(x^2-6x+9\right)+6\)

\(\Rightarrow A=\left(x-3\right)^2+6\)

Ta có: \(\left(x-3\right)^2+6\ge6\) với mọi x

Dấu ''='' xảy ra khi \(x=3\)

có làm mới có ăn nha em

GTNN của biểu thức A là 5

`A=x^4-6x^3+18x^2-6xy+y^2+2012`
`=x^4-6x^3+9x^2+9x^2-6xy+y^2+2012`
`=(x^2-x)^2+(3x-y)^2+2012>=2012`
Dấu "=" xảy ra khi:
$\begin{cases}x=x^2\\y=3x\end{cases}$
`<=>` $\left[ \begin{array}{l}\begin{cases}x=0\\y=3x=0\\\end{cases}\\\begin{cases}x=1\\y=3x=3\\\end{cases}\end{array} \right.$
Vậy `min_A=2012<=>` $\left[ \begin{array}{l}x=y=0\\\begin{cases}x=1\\y=3\end{cases}\end{array} \right.$

\(A=\sqrt{x-2}+\sqrt{4-x}\ge\sqrt{x-2+4-x}=\sqrt{2}\)

\(A_{min}=\sqrt{2}\) khi \(\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

\(y=4x^2+\dfrac{9}{x^2}-3\ge2\sqrt{\dfrac{36x^2}{x^2}}-3=9\)

\(y_{min}=9\) khi \(x^2=\dfrac{3}{2}\)

\(P=\dfrac{x-1}{4}+\dfrac{1}{x-1}+\dfrac{1}{4}\ge2\sqrt{\dfrac{x-1}{4\left(x-1\right)}}+\dfrac{1}{4}=\dfrac{5}{4}\)

\(P_{min}=\dfrac{5}{4}\) khi \(x=\dfrac{3}{2}\)

`A=x^2+6x+y^2+4y+15`

`=(x^2+6x+9)+(y^2+4y+4)+2`

`=(x+3)^2+(y+2)^2+2`

Vì `(x+3)^2+(y+2)^2 >=0 forall x,y`

`=>A_(min)=2 <=> x=-3; y=-2`.

Ta có: \(A=x^2+6x+y^2+4y+15\)

\(=x^2+6x+9+y^2+4y+4+2\)

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

Dấu '=' xảy ra khi (x,y)=(-3;-2)