CMR:[tex]\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}[/tex] với [tex]a\geq \frac{1}{8}[/tex] là số tự nhiên
Đặt [tex]x= \sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}+\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}[/tex]
[tex]\Rightarrow x^{3}=a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}+a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}+3x \left (\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}} \cdot \sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}} \right )[/tex]
[tex]=2a+3x\left ( \sqrt[3]{a^{2}-\frac{(a+1)^{2}(8a-1)}{27}} \right )[/tex]
[tex]=2a+3x\cdot \sqrt[3]{\frac{1-6a+12a^{2}-8a^{3}}{27}}[/tex]
[tex]=2a+3x\cdot \frac{1-2a}{3}=2a+x-2ax[/tex]
[tex]\Rightarrow x^{3}-x+2ax-2a=0[/tex]
[tex]\Leftrightarrow (x-1)(x^2+x+2a)=0[/tex]
Nếu [tex]a=\frac{1}{8}[/tex] thì $x=1$
Nếu $a > \frac{1}{8}$ thì $x^2+x+2a >x^2+x+\frac{1}{4}=(x+\frac{1}{2})^2 \geq 0$
$\Rightarrow x-1=0 \Rightarrow x=1$
Như vậy, với $a \geq \frac{1}{8}$ thì [tex]x=1 \in \mathbb{N}[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
