Question

determine o resultado
$\sqrt[3]{548} \times \frac{12}{4} \div 6 = {x}^{2} + 52x + 4x {}^{2}$

me ajudem
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Answer 1

Explicação passo-a-passo:

$\sqrt[3]{548} \times \frac{12}{4} \div 6 = {x}^{2} + 52x + {4x}^{2} \\ \\ \sqrt[3]{548} \times 3 \div 6 = {5x}^{2} + 52x \\ \\ \frac{ \sqrt[3]{548} \times 3 }{6} = {5x}^{2} + 52x \\ \\ \frac{ \sqrt[3]{548} }{2} = {5x}^{2} + 52x \\ \\ \sqrt[3]{548} = 10 {x}^{2} + 104x \\ \\ \sqrt[3]{548} - 10 {x}^{2} + 104x = 0 \\ \\ - 10 {x}^{2} - 104 x + \sqrt[3]{548} = 0 \\ \\ {10x}^{2} + 104x - \sqrt[3]{548} = 0 \\ \\ x = \frac{ - 104 ± \sqrt{ {104}^{2} - 4 \times 10 \times ( - \sqrt[3]{548}) } }{2 \times 10} \\ \\ x = \frac{ - 104 ± \sqrt{10816 + 40 \sqrt[3]{548} } }{20} \\ \\ x = \frac{ - 104 ±2 \sqrt{2704 + 10 \sqrt[3]{548} } }{20} \\ \\ x = \frac{ - 104 + 2 \sqrt{2704 + 10 \sqrt[3]{548} } }{20} \\ x = \frac{ - 104 - 2 \sqrt{2704 + 10 \sqrt[3]{548} } }{20} \\ \\ x_{1} = \frac{ - 52 + \sqrt{2704 + 10 \sqrt[3]{548} } }{10} \\ x_{2} = \frac{ - 52 - \sqrt{2704 + 10 \sqrt[3]{548} } }{10}$

• Espero ter ajudado.