\(A=2\left(x^2-2xy+y^2\right)+\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{8067}{4}\)

\(A=2\left(x-y\right)^2+\left(x-\dfrac{3}{4}\right)^2+\dfrac{8067}{4}\ge\dfrac{8067}{4}\)

\(A_{min}=\dfrac{8067}{4}\) khi \(x=y=\dfrac{3}{2}\)

\(A=x^2-x=\left(x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}\right)-\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Dấu "=" xảy ra khi \(x=\dfrac{1}{2}\) 

Vậy \(A_{min}=-\dfrac{1}{4}\)

A= x^2-x

A= (x-1/2)^2-1/4

ta thấy (x-1/2)^2\(\ge\)0

=>(x-1/2)^2-1/4\(\ge\)-1/4

hay A\(\ge\)-1/4

vậy \(A_{min}\)=-1/4<=>x=1/2

lập bảng xét dấu đi

\(A=-3x^2-5\left|y-1\right|+3\le3\)

Dấu ''='' xảy ra khi x = 0 ; y = 1

THAM KHẢO:

Dấu ''='' xảy ra khi x = 0 ; y = 1

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

1, Ta có: \(A=3x^2+8x+9=3\left(x^2+\frac{8}{3}x+3\right)=3\left(x^2+\frac{8}{3}x+\frac{16}{9}+\frac{11}{9}\right)\)

\(=3\left(x+\frac{4}{3}\right)^2+\frac{11}{3}\ge\frac{11}{3}\forall x\)

=> Min A = 11/3 tại x = -4/3

2, Ta có: \(A=-2x^2+6x+3=-2\left(x^2-3x-\frac{3}{2}\right)=-2\left(x^2-3x+\frac{9}{4}-\frac{15}{4}\right)\)

\(=-2\left(x-\frac{3}{2}\right)^2+\frac{15}{2}\le\frac{15}{2}\forall x\)

=> Max A = 15/2 tại x = 3/2

=.= hk tốt!!

Cảm ơn

\(A=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2\ge2\\ A_{min}=2\Leftrightarrow x=3\\ B=2\left(x^2-10x+25\right)+51=2\left(x-5\right)^2+51\ge51\\ B_{min}=51\Leftrightarrow x=5\\ C=\left[\left(x^2-4xy+4y^2\right)+10\left(x-2y\right)+25\right]+\left(y^2-2y+1\right)+2\\ C=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\\ C_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x-2y+5=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y-5=2-5=-3\\y=1\end{matrix}\right.\)

a) \(A=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2\ge2\)

\(minA=2\Leftrightarrow x=3\)

b) \(B=2\left(x^2-10x+25\right)+51=2\left(x-5\right)^2+51\ge51\)

\(minB=51\Leftrightarrow x=5\)

c) \(C=\left[x^2-2x\left(2y-5\right)+\left(2y-5\right)^2\right]+\left(y^2-2y+1\right)+2=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

\(minC=2\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\end{matrix}\right.\)

a)-19

b)22