`.a` \(A=\left(x^2-3x+1\right)\left(x^2-3x-1\right)\)
\(A=\left(x^2-3x\right)^2-1^2\)
\(A=\left(x^2-3x\right)^2-1\ge-1\)
Dấu "=" xảy ra \(\Leftrightarrow x^2-3x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy \(Min_A=-1\) khi \(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
`b.`\(B=\left(x-1\right)\left(x+5\right)\left(x^2+4x+5\right)\)
\(B=\left(x^2+4x-5\right)\left(x^2+4x+5\right)\)
\(B=\left(x^2+4x\right)^2-5^2\)
\(B=\left(x^2+4x\right)^2-25\ge-25\)
Dấu "=" xảy ra \(\Leftrightarrow x^2+4x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
Vậy \(Min_B=-25\) khi \(\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
a) \(A=\left(x^2-3x+1\right)\left(x^2-3x-1\right)\)
\(=\left(x^2-3x\right)^2-1\ge-1\)
Dấu bằng xảy ra \(\Leftrightarrow x^2-3x=0\Leftrightarrow x\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy \(MinA=-1\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
b) \(B=\left(x-1\right)\left(x+5\right)\left(x^2+4x+5\right)\)
\(=\left(x^2+4x-5\right)\left(x^2+4x+5\right)\)
\(=\left(x^2+4x\right)^2-25\)
Dấu bằng xảy ra \(\Leftrightarrow x^2+4x=0\Leftrightarrow x\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
Vậy \(MinB=-25\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
