a) \(A=2x^2+8x+12=2x^2+8x+8+4\)

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

Vì \(\left(x+2\right)^2\ge0\forall x\)\(\Rightarrow2\left(x+2\right)^2\ge0\forall x\)

\(\Rightarrow2\left(x+2\right)^2+4\ge4\forall x\)\(\Rightarrow A\ge4\)

Dấu " = " xảy ra \(\Leftrightarrow x+2=0\)\(\Leftrightarrow x=-2\)

Vậy \(A_{min}=4\)\(\Leftrightarrow x=-2\)

b) \(B=x^2-6x+30=x^2-6x+9+21=\left(x-3\right)^2+21\)

Vì \(\left(x-3\right)^2\ge0\forall x\)\(\Rightarrow\left(x-3\right)^2+21\ge21\forall x\)

\(\Rightarrow B\ge21\)

Dấu " = " xảy ra \(\Leftrightarrow x-3=0\)\(\Leftrightarrow x=3\)

Vậy \(B_{min}=21\)\(\Leftrightarrow x=3\)

\(x\left(x^2-y\right)-x^2\left(x-y\right)+1817\)

\(=x^3-xy-\left(x^3-x^2y\right)+1817\)

\(=x^3-xy-x^3+x^2y+1817\)

\(=xy\left(x-1\right)+1817\)

Thế \(x=-1\)và \(y=100\)vào biểu thức sau khi rút gọn ta được: 

\(\left(-1\right).100.\left[\left(-1\right)-1\right]+1817=-100.\left(-2\right)+1817=2017\)

a. x² - 2x - 3 = 0

<=> ( x² + x ) - ( 3x + 3 ) = 0

<=> x ( x + 1 ) - 3 ( x + 1 ) = 0

<=> ( x - 3 ) ( x + 1 ) = 0

<=> x = 3 hoặc x = - 1

b. 2x² - 3 + 5x = 0

<=> 2 ( x² + 5/2x - 3/2 ) = 0

<=> ( x² + 5/2x + 25/16 ) - 49/16 = 0

<=> ( x + 5/4 )² = 49/16

<=>\(\orbr{\begin{cases}x+\frac{5}{4}=\frac{7}{4}\\x+\frac{5}{4}=-\frac{7}{4}\end{cases}}\)<=>\(\orbr{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}\)

a) \(x^2-2x-3=0\)

\(x^2-2x+1-4=0\)

\(\left(x-1\right)^2-4=0\)

\(\left(x-1-2\right)\left(x-1+2\right)=0\)

\(\left(x-3\right)\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

b) \(2x^2-3+5x=0\)

\(2x^2-3+6x-x=0\)

\(\left(2x^2+6x\right)-\left(3+x\right)=0\)

\(2x\left(x+3\right)-\left(3+x\right)=0\)

\(\left(x+3\right)\left(2x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\2x-1=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-3\\x=\frac{1}{2}\end{cases}}\)