\(a)\)

\(\begin{array}{l}
A=x^2-x+3\\
A=\left(x^2-x+\frac{1}{4}\right)+\dfrac{11}{4}\\
A=\left(x-\frac{1}{2}\right)^2+\dfrac{11}{4}\\
\text{Nhan xet: }\left(x-\dfrac{1}{2}\right)^2\ge 0,\forall x\\
\to A=\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge \dfrac{11}{4},\forall x
\end{array}\)

Vậy \(A_{min}=\dfrac{11}{4}\leftrightarrow x=\dfrac{1}{2}\)

\(b)\)

\(\begin{array}{l}
B=x^2+y^2-x+4y+5\\
B=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+4y+4\right)+\frac{3}{4}\\
B=\left(x-\dfrac{1}{2}\right)^2+\left(y+2\right)^2+\frac{3}{4}\\
\text{Nhan xet: }\left(x-\dfrac{1}{2}\right)^2\ge 0,\forall x;\left(y+2\right)^2\ge 0,\forall y\\
\to B=\left(x-\dfrac{1}{2}\right)^2+\left(y+2\right)^2+\dfrac{3}{4}\ge \dfrac{3}{4},\forall x,y
\end{array}\)

Vậy \(B_{min}=\dfrac{3}{4}\leftrightarrow \left(x;y\right)=\left(\dfrac{1}{2};-2\right)\)

\(c)\)

\(\begin{array}{l}
C=2x^2+4y^2+4xy-3x-1\\
C=\left(x^2+4xy+4y^2\right)+\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{13}{4}\\
C=\left(x+2y\right)^2+\left(x-\dfrac{3}{2}\right)^2-\dfrac{13}{4}\\
\text{Nhan xet: }\left(x+2y\right)^2\ge 0;\left(x-\dfrac{3}{2}\right)^2\ge 0,\forall x,y\\
\to C=\left(x+2y\right)^2+\left(x-\dfrac{3}{2}\right)^2-\dfrac{13}{4}\ge -\dfrac{13}{4},\forall x,y
\end{array}\)

Vậy \(C_{min}=-\dfrac{13}{4}\leftrightarrow \begin{cases} x+2y=0\\ x-\dfrac{3}{2}=0 \end{cases}\to\begin{cases} x=\dfrac{3}{2}\\ y=-\dfrac{3}{4}\end{cases}\)

Đáp án:

`a,A_(min)=11/4 <=> x=1/2`

`b,B_(min)=3/4 <=> (x,y)=(1/2 ; -2}`

`c,C_(min)=-5/4 <=> (x,y)=(3/2 , -3/4)`

Giải thích các bước giải:

 `a, A= x^2 -x+3`

`=(x^2 -x +1/4 )+ 11/4`

`=[x^2 - 2. x . 1/2 + (1/2)^2] + 11/4`

`=(x- 1/2)^2 + 11/4`

Vì `(x- 1/2)^2 >= 0 forall => (x- 1/2)^2 + 11/4 >= 0+ 11/4 = 11/4 forall x`

Dấu "=" xảy ra `<=> x-1/2 =0 <=> x=1/2`

Vậy `A_(min)=11/4 <=> x=1/2`

`b, x^2 +y^2 - x +4y+5`

`=(x^2 -x + 1/4)+(y^2 + 4y + 4) + 3/4`

`=[x^2 - 2 .1/2 .x + (1/2)^2] + (y^2 +2.2y+2^2)+ 3/4`

`=(x- 1/2)^2 + (y+2)^2 + 3/4`

Vì `(x-1/2)^2 >= 0 forall x ; (y+2)^2 >= 0 forall y`

`=> (x- 1/2)^2 + (y+2)^2 +3/4 >= 0+0+ 3/4 = 3/4 forall x,y`

Dấu "=" xảy ra `<=>` $\begin{cases} x-\dfrac{1}{2}=0\\y+2=0 \end{cases}$

`<=>` $\begin{cases} x=\dfrac{1}{2}\\y=-2 \end{cases}$

Vậy `B_(min)=3/4 <=> (x,y)=(1/2 , -2}`

`c, C= 2x^2 +4y^2 +4xy-3x-1`

`=(x^2 +4xy+4y^2)+(x^2 -3x+9/4)-5/4`

`=[x^2 +2.x.2y+(2y)^2]+[x- 2 . 3/2 . x + (3/2)^2] -5/4`

`=(x+2y)^2 + (x- 3/2)^2 - 5/4`

Vì `(x+2y)^2 >=0 forall x, y ; (x- 3/2)^2 >= 0 forall x`

`=> (x+2y)^2 + (x- 3/2)^2 - 5/4 >= 0+0 - 5/4 = -5/4`

Dấu "=' xảy ra `<=>` $\begin{cases} x+2y=0=0\\x-3\dfrac{3}{2}=0 \end{cases}$

`<=>` $\begin{cases} \dfrac{3}{2}+2y=0\\x=\dfrac{3}{2} \end{cases}$

`<=>` $\begin{cases} 2y=-\dfrac{3}{2}\\x=\dfrac{3}{2} \end{cases}$

`<=>` $\begin{cases} y=-\dfrac{3}{4}\\x=\dfrac{3}{2} \end{cases}$

Vậy `C_(min)=-5/4 <=> (x,y)=(3/2 , -3/4)`