Question

Ayudenme pls.... :,)​

Answer 1

Verified answer

Respuesta:

a) [tex]18 + \sqrt_{3}(-3)[/tex]; b) 62; c) x = -9 ; d) x = -19

Explicación paso a paso:

a)  [tex] \sqrt[3]{-3} \cdot \sqrt[3]{27}+(-4)^{7}:(-4)^{5}-\left[(-14)^{3}\right]^{0}+\sqrt{\sqrt{81}} = 3 \sqrt[3]{-3} + 4^2 -1 +3 = 18 + 3 \sqrt[3]{-3} [/tex]

b)  [tex] \sqrt{\sqrt[3]{64}}+(-2)^{2} \cdot(-2)^{4}-\left[(-11)^{2}\right]^{0}+\sqrt[3]{-81}: \sqrt[3]{3} = \sqrt[6]{2^6} + 2^6 -1 +\sqrt[3]{-27} = 2 + 64 -1 -3 = 62 [/tex]

c)  [tex] 3x + 21 + 6x = -60 \Rightarrow 9x = -81 \Rightarrow x = -9  [/tex]

Comprobación: 3(-9+7) + 6(-9) = 3 (-2) + 6(-9) = -6 - 54 = -60 = -120/2

d)  [tex] 2(x-8) - 5 \sqrt{4} = 4x + 12 \Rightarrow 2x - 16 - 10 = 4x + 12 \Rightarrow  [/tex]

[tex] \Rightarrow 2x - 26 = 4x + 12  \Rightarrow -38 = 2x  \Rightarrow x = -19 [/tex]

Comprobación:

4(-19) + 12 = -76 + 12 = -64

2(-19-8) - 10 = -54 -10 = -64