\(C=\dfrac{x^2+8}{x^2-2}\)
\(x^2\ge0\Leftrightarrow\left\{{}\begin{matrix}x^2+8\ge8\\x^2-2\ge-2\end{matrix}\right.\)
Mà \(C\) nhỏ nhất nên: \(x^2-2\) phải lớn nhất
\(x^2+8>0\Leftrightarrow x^2-2< 0\) ( để C nhỏ nhất)
\(\Leftrightarrow x^2-2=-1\Rightarrow x^2=1\)
\(min_C=\dfrac{1+8}{1-2}=-9\) Xảy ra khi
\(x^2=1\Leftrightarrow x=\pm1\)
2)
\(A=x^4+3x^2\)
\(A=x^4+3x^2+\dfrac{9}{4}-\dfrac{9}{4}\)
\(A=\left(x^2+\dfrac{3}{2}\right)^2-\dfrac{9}{4}\)
\(x^2\ge0\Rightarrow x^2+\dfrac{3}{2}\ge\dfrac{3}{2}\Leftrightarrow\left(x^2+\dfrac{3}{2}\right)^2\ge\dfrac{9}{4}\)
\(A=\left(x^2+\dfrac{3}{2}\right)^2-\dfrac{9}{4}\ge0\)
Dấu "=" xảy ra khi:
\(x^2=0\Leftrightarrow x=0\)
\(B=\left(x^4+5\right)^2+2\)
\(x^4\ge0\Leftrightarrow x^4+5\ge5\Leftrightarrow\left(x^4+5\right)^2\ge25\)
\(B=\left(x^4+5\right)^2+2\ge27\)
Dấu "=" xảy ra khi:
\(x^4=0\Leftrightarrow x=0\)