\(C=\left(x-1\right)\left(x-3\right)\left(x+5\right)\left(x+7\right)=\left[\left(x+5\right)\left(x-1\right)\right]\left[\left(x+7\right)\left(x-3\right)\right]=\left(x^2+4x-5\right)\left(x^2+4x-21\right)\)
Đặt \(t=x^2+4x-5,\) khi đó ta có:
\(C=t\left(t-16\right)=t^2-16t=\left(t^2-2.8.t+8^2\right)-64=\left(t-8\right)^2-64\ge-64\)
Đẳng thức xảy ra \(\Leftrightarrow t-8=0\Leftrightarrow t=8\Leftrightarrow x^2+4x-5=8\Leftrightarrow x^2+4x-13=0\)
Vậy \(MinC=-64\Leftrightarrow x^2+4x-13=0\)
\(E=x^4+\left(3-x\right)^2=x^4+x^2-6x+9=\left(x^4-2x^2+1\right)+\left(3x^2-6x+3\right)+5=\left(x^2-1\right)^2+3\left(x-1\right)^2+5\ge5\)
Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x^2-1=0\\x-1=0\end{matrix}\right.\Leftrightarrow x=1\)
Vậy \(MinE=5\Leftrightarrow x=1\)
