Đặt \(C=\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)

\(=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)

\(=\left|2x-1\right|+\left|2x-3\right|\)

\(=\left|2x-1\right|+\left|3-2x\right|\)

\(\ge\left|\left(2x-1\right)+\left(3-2x\right)\right|=\left|2\right|=2\)

Vậy \(C_{min}=2\)

#)Giải :

\(\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)

\(=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)

\(=\left|2x-1\right|+\left|2x-3\right|\)

\(=\left|2x-1\right|+\left|3-2x\right|\ge\left|2x-1+3-2x\right|=2\)

Dấu ''='' xảy ra khi x = 1

\(\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)

\(=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)

\(=|2x-1|+|2x-3|\)

\(=|2x-1|+|3-2x|\ge|2x-1+3-2x|=2\)

Dấu"=" xảy ra \(\Leftrightarrow\left(2x-1\right)\left(3-2x\right)\ge0\Leftrightarrow\frac{1}{2}\le x\le\frac{3}{2}\)

Chỉ là góp ý:V

\(A=4x^2+y^2-12x+8y+28\)

\(=\left(4x^2-12x+9\right)+\left(y^2+8y+16\right)+3\)

\(=\left[\left(2x\right)^2+2.2x.3+3^2\right]+\left(y^2+2.y.4+4^2\right)+3\)

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

Ta có :

\(\left(2x+3\right)^2\ge0\forall x\)

\(\left(y+4\right)^2\ge0\forall x\)

\(\Rightarrow\left(2x+3\right)^2+\left(y+4\right)^2+3\ge3\forall x\)

Dấu = xảy ra khi

\(\left\{{}\begin{matrix}\left(2x+3\right)^2=0\\\left(y+4\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x+3=0\\y+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{2}\\y=-4\end{matrix}\right.\)

Vậy \(Min_A=3\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=-4\end{matrix}\right.\)

a) \(M=x^2+10x+28=\left(x^2+10+25\right)+3=\left(x+5\right)^2+3\ge3\)

\(minM=3\Leftrightarrow x=-3\)

b) \(P=4x^2-12x+10=\left(4x^2-12x+9\right)+1=\left(2x-3\right)^2+1\ge1\)

\(minP=1\Leftrightarrow x=\dfrac{3}{2}\)

Lời giải:

a) 

$A=4x^2+4x+11=(4x^2+4x+1)+10=(2x+1)^2+10\geq 10$

Vậy $A_{\min}=10$. Giá trị này đạt tại $(2x+1)^2=0$

$\Leftrightarrow x=-\frac{1}{2}$

b) 

$C=x^2-2x+y^2-4y+7=(x^2-2x+1)+(y^2-4y+4)+2$

$=(x-1)^2+(y-2)^2+2\geq 2$

Vậy $C_{\min}=2$. Giá trị này đạt tại $(x-1)^2=(y-2)^2=0$

$\Leftrightarrow x=1; y=2$

1:

=x^2-6x+9-4=(x-3)^2-4>=-4

Dấu = xảy ra khi x=3

3: =-y^2-4y-4+13

=-(y+2)^2+13<=13

Dấu = xảy ra khi y=-2

4: D=x^2-8>=-8

Dấu = xảy ra khi x=0

\(A=x^2-4x+20=x^2-4x+4+16=\left(x-2\right)^2+16\)

Do \(\left(x-2\right)^2\ge0\)

\(\Rightarrow\left(x-2\right)^2+16\ge16\)

\(\Rightarrow Min\left(A\right)=16\)

\(B=x^2-3x+7=x^2-3x+\dfrac{9}{4}-\dfrac{9}{4}+7=\left(x-\dfrac{3}{2}\right)^2+\dfrac{19}{4}\)

Do \(\left(x-\dfrac{3}{2}\right)^2\ge0\)

\(\Rightarrow\left(x-\dfrac{3}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}\)

\(\Rightarrow Min\left(B\right)=\dfrac{19}{4}\)

\(C=-x^2-10x+70=-\left(x^2+10x+25\right)+25+70=-\left(x-5\right)^2+95\)

Do \(-\left(x-5\right)^2\le0\)

\(\Rightarrow-\left(x-5\right)^2+95\le95\)

\(\Rightarrow Max\left(C\right)=95\)

\(D=-4x^2+12x+1=-\left(4x^2-12x+9\right)+9+1=-\left(2x-3\right)^2+10\)

Do \(-\left(2x-3\right)^2\le0\)

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

\(\Rightarrow Max\left(D\right)=10\)

C = 2x² + 2y² + 26 + 12x - 8y

C = (2x² + 12x + 18) + (2y² - 8y + 8) 

C = 2(x² + 6x + 9) + 2(y² - 4y + 4)

C = 2(x + 3)² + 2(y - 2)² \(\ge\)0 với mọi x;y

Dấu "=" xảy ra <=> x + 3 = 0 và y - 2 = 0

<=> x = -3 và y = 2

Vậy MinC = 0 khi x = -3 và y = 2

\(C=2\left(x^2+6x+9\right)+2\left(y^2-4y+4\right)=2\left(x+3\right)^2+2\left(y-2\right)^2\ge0\)

Vậy MIN C=0 khi và chỉ khi x+3=y-2=0 suy ra x=-3;y=2

C = 2x² + 2y² + 26 + 12x - 8y

C = ( 2x² + 12x + 18 ) + ( 2y² - 8y + 8 )

C = 2( x² + 6x + 9 ) + 2( y² - 4y + 4 )

C = 2( x + 3 )² + 2( y - 2 )²

\(\hept{\begin{cases}2\left(x+3\right)^2\ge0\forall x\\2\left(y-2\right)^2\ge0\forall y\end{cases}}\Rightarrow2\left(x+3\right)^2+2\left(y-2\right)^2\ge0\forall x,y\)

Đẳng thức xảy ra <=> \(\hept{\begin{cases}x+3=0\\y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=2\end{cases}}\)

Vậy C_Min = 0 , đạt được khi x = -3 , y = 2