Question

aplicar propiedades para resolver (sin elevar).
$a) \sqrt{ - 1} \times \sqrt{ - 9}$
$b)( - 3 { \times }^{2} {)}^{3}$
Hallar x :
$a) {x}^{3} + 30 = 3$
$b) \sqrt[4]{2} x - 2 = 2$
$c) \sqrt{3} x - 4 = 7$

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Answer 1

Respuesta:

$( - 3 {x}^{2} {)}^{3} =$

$( - 3 {)}^{3} \times ( {x}^{2} {)}^{3} =$

$- 27 {x}^{6}$

Hallar x :

${x}^{3} + 30 = 3$

${x}^{3} = 3 - 30$

${x}^{3 } = - 27$

${x}^{3 } = ( - 3 {)}^{3}$

$x = - 3$

$\sqrt[4]{2x} - 2 = 2$

$\sqrt[4]{2x} = 2 + 2$

$\sqrt[4]{2x} = 4$

$2x = 256$

$x = 128$

$\sqrt{3x} - 4 = 7$

$\sqrt{3x } = 7 + 4$

$\sqrt{3x} = 11$

$3x = 121$

$x = 121 \div 3$

$x = 40$