Muốn lý do ak giống nhau thôi

\(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

\(\Leftrightarrow A=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)

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

\(\Leftrightarrow A=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(\Leftrightarrow A=\left(x^2+5x\right)^2-36\ge-36\forall x\)

Dấu " = " xảy ra

\(\Leftrightarrow x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

Vậy GTNN của A là : \(-36\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

\(A=36-3x+\dfrac{1}{2}x^2=\dfrac{1}{2}\left(x^2-6x+72\right)\)

\(=\dfrac{1}{2}\left[\left(x^2-6x+9\right)+63\right]=\dfrac{1}{2}\left[\left(x-3\right)^2+63\right]\)

Có: \(\left(x-3\right)^2\ge0\forall x\Rightarrow\left(x-3\right)^2+63\ge63\)

\(\dfrac{1}{2}\left[\left(x-3\right)^2+63\right]\ge\dfrac{1}{2}\cdot63=\dfrac{63}{2}\)

Dấu ''='' xảy ra khi x = 3

Vậy \(MIN_A=\dfrac{63}{2}\Leftrightarrow x=3\)

a)

\(\dfrac{P}{Q}=\dfrac{R}{S}\Rightarrow PS=QR\)

\(\Leftrightarrow PS+QS=QR+QS\)

\(\Leftrightarrow S\left(P+Q\right)=Q\left(R+S\right)\)

điều kiện Q,s khác 0 => chia hau vế cho QS

\(\Leftrightarrow\dfrac{S\left(P+Q\right)}{QS}=\dfrac{Q\left(R+S\right)}{QS}\Leftrightarrow\dfrac{\left(P+Q\right)}{Q}=\dfrac{\left(R+S\right)}{S}\) đpcm

\(B=-3x^2+x+1\)

\(B=-3\left(x^2-\dfrac{1}{3}x-\dfrac{1}{3}\right)\)

\(B=-3\left[\left(x^2-2\cdot x\cdot\dfrac{1}{6}+\dfrac{1}{36}\right)-\dfrac{13}{36}\right]\)

\(B=-3\left(x-\dfrac{1}{6}\right)^2+\dfrac{13}{12}\)\(\le\dfrac{13}{12}\forall x\)

\(B=\dfrac{13}{12}\Leftrightarrow-3\left(x-\dfrac{1}{6}\right)^2=0\Leftrightarrow x=\dfrac{1}{6}\)

Vậy Max B = 13/12 <=> x = 1/6