\(A=x^2-5x+1=x^2+5x-\dfrac{21}{4}+\dfrac{25}{4}=\left(x^2+5x+\dfrac{25}{4}\right)-\dfrac{21}{4}\)

                                                                \(=\left[x^2+2.\dfrac{5}{2}.x+\dfrac{5}{2}^2\right]-\dfrac{21}{4}\)

                                                                \(=\left(x+\dfrac{5}{2}\right)^2-\dfrac{21}{4}\)

Do \(\left(x+\dfrac{5}{2}\right)^2\ge0\forall x\Rightarrow\left(x+\dfrac{5}{2}\right)^2-\dfrac{21}{4}\ge-\dfrac{21}{4}\forall x\)

Dấu '' = '' xảy ra khi \(x=-\dfrac{5}{2}\)

Vậy \(min_A=-\dfrac{21}{4}\) khi \(x=-\dfrac{5}{2}\)

Bấm máy.

\(P=\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)

\(=\sqrt{13+30\sqrt{2+\sqrt{8+4\sqrt{2}+1}}}\)

\(=\sqrt{13+30\sqrt{2+\sqrt{\left(\sqrt{8}+1\right)^2}}}\)

\(=\sqrt{13+30\sqrt{2+\sqrt{8}+1}}\)

\(=\sqrt{13+30\sqrt{3+\sqrt{8}}}\)

\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\)

\(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)

\(=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)

\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)

\(=\sqrt{13+30\sqrt{2}+30}\)

\(=\sqrt{43+30\sqrt{2}}\)

\(=\sqrt{\left(5+3\sqrt{2}\right)^2}\)

\(=5+3\sqrt{2}\)

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