Đáp án:
a) $ x = \dfrac{19}{30}$
b) $ x = 0$
Giải thích các bước giải:
Áp dụng HĐT $: (a + b)³ = a³ + b³ + 3ab(a + b)$
a) $PT ⇔ (\sqrt[3]{2x - 1} + \sqrt[3]{3x - 1})³ = (\sqrt[3]{5x + 1})³$
$ ⇔ (2x - 1) + (3x - 1) + 3\sqrt[3]{2x - 1}.\sqrt[3]{3x - 1}.(\sqrt[3]{2x - 1} + \sqrt[3]{3x - 1}) = 5x + 1$
$ ⇔ \sqrt[3]{2x - 1}.\sqrt[3]{3x - 1}.\sqrt[3]{5x + 1} = 1$
$ ⇔ (2x - 1)(3x - 1)(5x + 1) = 1$
$ ⇔ x²(30x - 19) = 0 $
$ ⇔ x = \dfrac{19}{30} (TM); x = 0 (ko TM) $
b) $PT ⇔ (\sqrt[3]{x + 1} + \sqrt[3]{x - 1})³ = (\sqrt[3]{5x})³$
$ ⇔ (x + 1) + (x - 1) + 3\sqrt[3]{x + 1}.\sqrt[3]{x - 1}.(\sqrt[3]{x + 1} + \sqrt[3]{x - 1}) = 5x $
$ ⇔ \sqrt[3]{x² - 1}.\sqrt[3]{5x} = x$
$ ⇔ 5x(x² - 1) = x³$
$ ⇔ x(4x² - 5) = 0 $
$ ⇔ x = 0 (TM) ; x = ± \dfrac{\sqrt{5}}{2} (ko TM)$
