Question

gente me ajudar, por favor​

Answer 1

Explicação passo-a-passo:

a)

$\sf{\sqrt[3]{x^2 - 7x} = 2}$

$\sf{(\sqrt[3]{x^2 - 7x})^3 = (2)^3}$

$\sf{x^2 - 7x = 8}$

$\sf{x^2 - 7x - 8 = 0}$

$\sf{\Delta = b^2 - 4ac}$

$\sf{\Delta = (-7)^2 - 4 \times 1 \times (-8)}$

$\sf{\Delta = 49 + 32}$

$\sf{\Delta = 81}$

$\sf{x = \dfrac{- b \pm \sqrt{\Delta}}{2a}}$

$\sf{x = \dfrac{- (-7) \pm \sqrt{81}}{2 \times 1}}$

$\sf{x = \dfrac{7 \pm 9}{2}}$

$•~~\sf{x' = \dfrac{7 + 9}{2} = \dfrac{16}{2} = 8}$

$•~~\sf{x'' = \dfrac{7 - 9}{2} = - \dfrac{2}{2} = - 1}$

$\boxed{\sf{S = \left\{- 1~~;~~8\right\}}}$

b)

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

$\sf{(\sqrt[4]{x^2 + x + 4})^4 = (2)^4}$

$\sf{x^2 + x + 4 = 16}$

$\sf{x^2 + x + 4 - 16 = 0}$

$\sf{x^2 + x - 12 = 0}$

$\sf{\Delta = b^2 - 4ac}$

$\sf{\Delta = 1^2 - 4 \times 1 \times (-12)}$

$\sf{\Delta = 1 + 48}$

$\sf{\Delta = 49}$

$\sf{x = \dfrac{- b \pm \sqrt{\Delta}}{2a}}$

$\sf{x = \dfrac{- 1 \pm \sqrt{49}}{2 \times 1}}$

$\sf{x = \dfrac{- 1 \pm 7}{2}}$

$•~~\sf{x' = \dfrac{- 1 + 7}{2} = \dfrac{6}{2} = 3}$

$•~~\sf{x'' = \dfrac{- 1 - 7}{2} = - \dfrac{8}{2} = - 4}$

$\boxed{\sf{S = \left\{- 4~~;~~3\right\}}}$

c)

$\sf{\sqrt{x^2 - 3x + 5} = \sqrt{2x - 1}}$

$\sf{(\sqrt{x^2 - 3x + 5})^2 = (\sqrt{2x - 1})^2}$

$\sf{x^2 - 3x + 5 = 2x - 1}$

$\sf{x^2 - 3x + 5 - 2x + 1 = 0}$

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

$\sf{\Delta = b^2 - 4ac}$

$\sf{\Delta = (-5)^2 - 4 \times 1 \times 6}$

$\sf{\Delta = 25 - 24}$

$\sf{\Delta = 1}$

$\sf{x = \dfrac{- b \pm \sqrt{\Delta}}{2a}}$

$\sf{x = \dfrac{- (-5) \pm \sqrt{1}}{2 \times 1}}$

$\sf{x = \dfrac{5 \pm 1}{2}}$

$•~~\sf{x' = \dfrac{5 + 1}{2} = \dfrac{6}{2} = 3}$

$•~~\sf{x'' = \dfrac{5 - 1}{2} = \dfrac{4}{2} = 2}$

$\boxed{\sf{S = \left\{2~~;~~3\right\}}}$

d)

$\sf{\sqrt{x^2 + 4x + 1} - \sqrt{x^2 + 17} = 0}$

$\sf{\sqrt{x^2 + 4x + 1} = \sqrt{x^2 + 17}}$

$\sf{(\sqrt{x^2 + 4x + 1})^2 = (\sqrt{x^2 + 17})^2}$

$\sf{x^2 + 4x + 1 = x^2 + 17}$

$\sf{x^2 + 4x + 1 - x^2 - 17 = 0}$

$\sf{4x - 16 = 0}$

$\sf{4x = 16}$

$\sf{x = \dfrac{16}{4}}$

$\sf{x = 4}$

$\boxed{\sf{S = \left\{4\right\}}}$