\(A=x^2-20x+101=\left(x-10\right)^2+1\ge1\)

\(minA=1\Leftrightarrow x=10\)

\(B=2x^2+40x-1=2\left(x+10\right)^2-201\ge-201\)

\(minB=-201\Leftrightarrow x=-10\)

\(C=x^2-4xy+5y^2-2y+28=\left(x^2-4xy+4y^2\right)+\left(y^2-2y+1\right)+27=\left(x-2y\right)^2+\left(y-1\right)^2+27\ge27\)

\(minC=27\Leftrightarrow\)\(\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)

\(D=\left(x-2\right)\left(x-5\right)\left(x^2-7x-10\right)=\left(x^2-7x+10\right)\left(x^2-7x+10\right)=\left(x^2-7x\right)^2-100\ge-100\)

\(minD=100\Leftrightarrow\)\(\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)

a: Ta có: \(A=x^2-20x+101\)

\(=x^2-20x+100+1\)

\(=\left(x-10\right)^2+1\ge1\forall x\)

Dấu '=' xảy ra khi x=10

b: ta có: \(B=2x^2+40x-1\)

\(=2\left(x^2+20x-\dfrac{1}{2}\right)\)

\(=2\left(x^2+20x+100-\dfrac{201}{2}\right)\)

\(=2\left(x+10\right)^2-201\ge-201\forall x\)

Dấu '=' xảy ra khi x=-10

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

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

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

Đáp án đúng : A

Dấu “=” xảy ra  ⇔ 2 a − 1 3 − 2 a ≥ 0 ⇔ 1 2 ≤ a ≤ 3 2

Vậy GTNN của B là 2 khi  1 2 ≤ a ≤ 3 2

a: Ta có: \(A=x^2-20x+101\)

\(=x^2-20x+100+1\)

\(=\left(x-10\right)^2+1\ge1\forall x\)

Dấu '=' xảy ra khi x=10

\(A=\left(x^2+4x+4\right)+3=\left(x+2\right)^2+3\ge3\)

\(A_{min}=3\) khi \(x=-2\)

\(B=\left(x^2-20x+100\right)+1=\left(x-10\right)^2+1\ge1\)

\(B_{min}=1\) khi \(x=10\)

\(C=\left(x^2+4y^2+25-4xy+10x-20y\right)+\left(y^2-2y+1\right)+2\)

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

\(C_{min}=2\) khi \(\left(x;y\right)=\left(-3;1\right)\)

`A=x^2-4x+y^2-8y+6`

`A=x^2-4x+4+y^2-8y+16-14`

`A=(x-2)^2+(y-4)^2-14`

VÌ `(x-2)^2+(y-4)^2>=0`

`=>(x-2)^2+(y-4)^2-14>=-14`

`=>A>=-14`

Dấu "=" xảy ra khi `x-2=0,y-4=0<=>{(x=2),(y=4):}`

a)-19

b)22

Bổ sung điều kiện: \(x,y>0\)

\(A=\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{xy}{x^2+y^2}\\ A=\dfrac{8}{9}\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\dfrac{1}{9}\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\dfrac{xy}{x^2+y^2}\\ A=\dfrac{8}{9}\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfrac{x^2+y^2}{9xy}+\dfrac{xy}{x^2+y^2}\right)\)

Áp dụng BĐT cosi:

\(A\ge\dfrac{8}{9}\cdot2\sqrt{\dfrac{xy}{xy}}+2\sqrt{\dfrac{xy\left(x^2+y^2\right)}{9xy\left(x^2+y^2\right)}}=\dfrac{16}{9}+\dfrac{2}{3}=\dfrac{22}{9}\)

Vậy \(A_{min}=\dfrac{22}{9}\Leftrightarrow x=y\)