Bài 4:
1)\(A=x^2-12x+7=\left(x^2-2.6x+36\right)-29\)\(=\left(x-6\right)^2-29\ge-29;\forall x\)
\(\Rightarrow A_{min}=-29\Leftrightarrow x-6=0\Leftrightarrow x=6\)
2)\(B=x^2+x+2=\left(x^2+2.\dfrac{1}{2}.x+\dfrac{1}{4}\right)+\dfrac{7}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\)\(\ge\dfrac{7}{4};\forall x\)
\(\Rightarrow B_{min}=\dfrac{7}{4}\Leftrightarrow x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{1}{2}\)
3)\(C=\dfrac{1-4x}{x^2}=\dfrac{4x^2-4x+1-4x^2}{x^2}\)\(=\dfrac{\left(2x-1\right)^2}{x^2}-4\ge-4;\forall x\)
\(\Rightarrow C_{min}=-4\Leftrightarrow2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)
4)\(D=x+y\)
Áp dụng bđt cosi có: \(D=x+y\ge2\sqrt{xy}=2\sqrt{25}=10\)
\(\Rightarrow D_{min}=10\Leftrightarrow\left\{{}\begin{matrix}x=y\\xy=25\end{matrix}\right.\)\(\Leftrightarrow x=y=5\)
5) \(E=x^3+y^3\)\(=\left(x+y\right)^3-3xy\left(x+y\right)=8-6xy\)
Có \(\left(x+y\right)^2\ge4xy\) \(\Leftrightarrow4\ge4xy\Leftrightarrow xy\le1\) \(\Rightarrow-xy\ge-1\)
\(\Rightarrow E=8-6xy=8+6.\left(-xy\right)\ge8+6.-1=2\)
\(\Rightarrow E_{min}=2\Leftrightarrow\left\{{}\begin{matrix}x=y\\x+y=2\end{matrix}\right.\)\(\Leftrightarrow x=y=1\)
6) \(F=x^4+\left(3-x\right)^2\)\(=x^4+x^2-6x+9=\left(x^4-2x^2+1\right)+\left(3x^2-6x+3\right)+5\)\(=\left(x^2-1\right)^2+3\left(x-1\right)^2+5\ge5;\forall x\)
\(\Rightarrow F_{min}=5\Leftrightarrow\left\{{}\begin{matrix}x^2-1=0\\x-1=0\end{matrix}\right.\)\(\Rightarrow x=1\)
