Question

resolva, no conjunto dos reais, a equação


30/³raiz 35-x³= x+³raiz 35-x³

URGENTE PFF​

Answer 1

Resposta:

$\textsf{Leia abaixo}$

Explicação passo a passo:

$\sf \dfrac{30}{x\:.\:\sqrt[\sf 3]{\sf 35 - x^3}} = x + \sqrt[\sf 3]{\sf 35 - x^3}$

$\sf x^2\:.\:\sqrt[\sf 3]{\sf 35 - x^3} + x\:.\:(\sqrt[\sf 3]{\sf 35 - x^3})^2 = 30$

$\sf{(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3}$

$\sf{a^2b + ab^2 = \dfrac{(a + b)^3 - (a^3 + b^3)}{3}}$

$\sf \dfrac{x^2\:.\:\sqrt[\sf 3]{\sf 35 - x^3} + x\:.\:(\sqrt[\sf 3]{\sf 35 - x^3})^2 - (x^3 + (\sqrt[\sf 3]{\sf 35 - x^3})^3)}{3} = 30$

$\sf x^2\:.\:\sqrt[\sf 3]{\sf 35 - x^3} + x\:.\:(\sqrt[\sf 3]{\sf 35 - x^3})^2 - (x^3 + 35 - x^3}) = 90$

$\sf (x + \sqrt[\sf 3]{\sf 35 - x^3})^3 - 35 = 90$

$\sf (x + \sqrt[\sf 3]{\sf 35 - x^3})^3 = 125$

$\sf x + \sqrt[\sf 3]{\sf 35 - x^3} = 5$

$\sf \sqrt[\sf 3]{\sf 35 - x^3} = 5 - x$

$\sf (\sqrt[\sf 3]{\sf 35 - x^3})^3 = (5 - x)^3$

$\sf 35 - x^3 = 5^3 - 3\:.\:5^2\:.\:x + 3\:.\:5\:.\:x^2 - x^3$

$\sf 35 = 125 - 75x + 15x^2$

$\sf 15x^2 -75x + 90 = 0$

$\sf x^2 -5x + 6 = 0$

$\sf a = 1 \Leftrightarrow b = -5 \Leftrightarrow c = 6$

$\sf x_1 \:.\:x_2 = \dfrac{c}{a} = \dfrac{6}{1}$

$\boxed{\boxed{\sf x_1 \:.\:x_2 = 6}}\leftarrow\textsf{letra A}$