Question

Me ajuda a resolver a questão 2 para mim​

Answer 1

a)

$x + \sqrt{20} = - \sqrt{45} \\ x = - \sqrt{45} - \sqrt{20} \\ x = - \sqrt{45 - 20} \\ x = - \sqrt{25} \\ x = - 5$

b)

$\sqrt{200} - 3x = \sqrt{72} \\ - 3x = \sqrt{72} - \sqrt{200} \\ 3x = - \sqrt{72} + \sqrt{200} \\ 3x = \sqrt{128} \\ x = \frac{ \sqrt{128} }{3} = \frac{8 \sqrt{2} }{3}$

c)

$67 + \sqrt{63} = x \sqrt{28} \\ x = \frac{67 + \sqrt{63} }{ \sqrt{28} } \\ \\ x = \frac{67 + \sqrt{63} }{ \sqrt{28} } \times \frac{ \sqrt{28} }{ \sqrt{28} } \\ \\ x = \frac{67 + \sqrt{1764} }{ \sqrt{784} } \\ \\ x = \frac{67 + 42}{28} = \frac{109}{28}$

d)

$2x + 2 \sqrt{10} = x - 3 \sqrt{90} \\ 2x - x = - 3 \sqrt{90} - 2 \sqrt{10} \\x = - 3 \times 3 \sqrt{10} - 2 \sqrt{10} \\ x = - 9 \sqrt{10} - 2 \sqrt{10} \\ x = - 11 \sqrt{10}$