Hãy tính : [tex]A = 2x^{3} + 2x^{2} + 1[/tex] với [tex]x = \frac{1}{3}.\left ( \sqrt[3]{\frac{23 + \sqrt{513}}{4}} + \sqrt[3]{\frac{23 - \sqrt{513}}{4}} - 1 \right )[/tex]
Đặt [tex]a=\sqrt[3]{\frac{23 + \sqrt{513}}{4}} + \sqrt[3]{\frac{23 - \sqrt{513}}{4}}\\\Leftrightarrow a^3=\frac{23 + \sqrt{513}}{4}+\frac{23 -\sqrt{513}}{4}+3.\sqrt[3]{\frac{23 + \sqrt{513}}{4}} . \sqrt[3]{\frac{23 - \sqrt{513}}{4}}.\left ( \sqrt[3]{\frac{23 + \sqrt{513}}{4}} + \sqrt[3]{\frac{23 - \sqrt{513}}{4}} \right )\\\Rightarrow a^3=\frac{23}{2}+3.a[/tex]
Ta có:
$x = \frac{1}{3}.\left ( \sqrt[3]{\frac{23 + \sqrt{513}}{4}} + \sqrt[3]{\frac{23 - \sqrt{513}}{4}} - 1 \right )\\\Leftrightarrow 3x+1=\sqrt[3]{\frac{23 + \sqrt{513}}{4}} + \sqrt[3]{\frac{23 - \sqrt{513}}{4}}\\\Leftrightarrow 3x+1=a$
Suy ra [tex](3x+1)^3=\frac{23}{2}+3.(3x+1)\\\Leftrightarrow 27x^3+27x^2+9x+1=\frac{23}{2}+9x+3\\\Leftrightarrow 27x^3+27x^2=\frac{27}{2}\\\Leftrightarrow 2x^3+2x^2=1\\\Leftrightarrow A=2x^3+2x^2+1=2[/tex]
Vậy $A=2$
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
