Question

a)Cho x=\(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}+1}\)

Tính P=(3\(x^3-9x^2-13)^{2015}\)

b)Tìm gtnn của A=\(\dfrac{x\sqrt{x}-6x+9\sqrt{x}}{4\left(\sqrt{x}-1\right)}.\left(\dfrac{3\sqrt{x}-5}{\sqrt{x}-3}-1\right)\)

Answer 1

a) tương tự : https://hoc24.vn/hoi-dap/question/650070.html

b) ta có : \(A=\dfrac{x\sqrt{x}-6x+9\sqrt{x}}{4\left(\sqrt{x}-1\right)}.\left(\dfrac{3\sqrt{x}-5}{\sqrt{x}-3}-1\right)\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)^2}{4\left(\sqrt{x}-1\right)}.\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}\right)=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2}=\dfrac{x-3\sqrt{x}}{2}\)

\(\Rightarrow x-3\sqrt{x}-2A=0\)

vì phương trình này luôn có nghiệm \(\Rightarrow\Delta\ge0\)

\(\Leftrightarrow3^2-4\left(-2A\right)=9+8A\ge0\Leftrightarrow A\ge\dfrac{-9}{8}\)

\(\Rightarrow\) GTNN của \(A=\dfrac{-9}{8}\) khi \(\sqrt{x}=\dfrac{-b}{2a}=\dfrac{3}{2}\)\(\Leftrightarrow x=\dfrac{9}{4}\)