Câu 1:
ta có: $x = \sqrt[3]{a+\dfrac{a+1}{3}\sqrt{\dfrac{8a-1}{3}}}+\sqrt[3]{a-\dfrac{a+1}{3}\sqrt{\dfrac{8a-1}{3}}}$
$\leftrightarrow x^3 = a+\dfrac{a+1}{3}\sqrt{\dfrac{8a-1}{3}}+a-\dfrac{a+1}{3}\sqrt{\dfrac{8a-1}{3}}+3\sqrt[3]{(a+\dfrac{a+1}{3}\sqrt{\dfrac{8a-1}{3}})(a-\dfrac{a+1}{3}\sqrt{\dfrac{8a-1}{3}})} [ \sqrt[3]{a+\dfrac{a+1}{3}\sqrt{\dfrac{8a-1}{3}}}+\sqrt[3]{a-\dfrac{a+1}{3}\sqrt{\dfrac{8a-1}{3}}} ]$
$\leftrightarrow x^3 = 2a+3\sqrt[3]{\dfrac{-(2a-1)^3}{27}}.x$
$\leftrightarrow x^3 = 2a-(2a-1)x$
$\leftrightarrow (x-1)(x^2+x+2a) = 0$
$\leftrightarrow \left[\begin{matrix} x = 1 \\ x^2+x+2a = 0 (1) \end{matrix}\right.$
(1) $\leftrightarrow x^2+x+2a = 0 \leftrightarrow (x+\dfrac{1}{2})^2+2a-\dfrac{1}{4} = 0$ (2)
Vì $x \in N$ và $a \ge \dfrac{1}{8}$ nên (2) vô nghiệm.
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