Question

alguém pra me ajudar por favorrr ​

Answer 1

Resposta:

a)

$3x^{2} = 0\\\\x^{2} = 0\\\\x = \sqrt{0} \\\\x = 0$

b)

$-9x^{2} = 0\\\\x^{2} = 0\\\\x = \sqrt{0} \\\\x = 0$

c)

$\frac{x4 . x^{2} }{7} = 0\\\\4x^{2} = 7 . 0\\\\x^{2} = \frac{0}{4}\\\\x = \sqrt{0}\\\\x = 0$

d)

$7x^{2} - 10 = 0\\\\7x^{2} = 10\\\\x^{2} = \frac{10}{7}\\\\x = \sqrt{\frac{10}{7} }$

e)

$x^{2} - 81 = 0\\\\x^{2} = 81\\\\x = \sqrt{81}\\\\x = 9$

f)

$x^{2} + 64 = 0\\\\x^{2} = -64\\\\x = \sqrt{-64}\\\\x = 8i$

g)

$\frac{4 . x^{2} }{3} - 3 = 0\\\\ \frac{4 . x^{2} }{3} = 3\\\\\frac{4 . x^{2} } = 3 . 3\\\\x^{2} = \frac{9}{4}\\\\x = \sqrt{\frac{9}{4} }\\\\x = \frac{3}{2}$

h)

$2x^{2} + 6x = 0\\\\delta = 6^{2} - (4 . 2 . 0)\\\\delta = 36\\\\x = \frac{-6 +/- \sqrt{36} }{2 .2 } \\\\x = \frac{-6 +/- 6}{4}\\\\x1 = \frac{-6 +6}{4} = 0\\\\x2 = \frac{-6 -6}{4} = \frac{-12}{4} = -3$

i)

$-x^{2} + x = 0\\\\delta = 1^{2} - (4 . -1 . 0)\\\\delta = 1\\\\x = \frac{-1 +/- \sqrt{1} }{2 . -1} \\\\x = \frac{-1 +/- 1}{-2}\\\\x1 = \frac{-1 - 1}{-2} = \frac{-2}{-2} = 1\\\\x2 = \frac{-1 + 1}{-2} = 0$

j)

$\frac{3x^{2} }{5} - x = 0\\\\delta = (-1)^{2} - (4 . \frac{3}{5} . 0)\\\\delta = 1\\\\x = \frac{-(-1) +/- \sqrt{1} }{2. \frac{3}{5} }\\\\x = \frac{1 +/- 1 }{\frac{6}{5} }\\\\x1 = \frac{1 - 1 }{\frac{6}{5} } = 0\\\\x2 = \frac{1 + 1 }{\frac{6}{5} } = \frac{2}{\frac{6}{5} } = \frac{2}{1} . \frac{5}{6} = \frac{10}{6} = \frac{5}{3}$

k)

$5x^{2} + \frac{2x}{5} = 0\\\\ delta = (\frac{2}{5} )^{2} - (4 . 5 . 0)\\\\delta = \frac{4}{25}\\\\x = \frac{-\frac{2}{5} +/- \sqrt{\frac{4}{25} } }{2 . 5}\\\\x = \frac{-\frac{2}{5} +/- \frac{2}{5} }{10} \\\\x1 = \frac{-\frac{2}{5} + \frac{2}{5} }{10} = 0 \\\\x2 = \frac{-\frac{2}{5} - \frac{2}{5} }{10} = \frac{-\frac{4}{5} }{10} = \frac{-4}{5} . \frac{1}{10} = \frac{-4}{50} = \frac{-2}{25}$

L)

$-\frac{6}{7} ^{2} + \frac{6x}{7} = 0\\\\delta = (\frac{6}{7}) ^{2} - (4 . -\frac{6}{7} . 0)\\\\delta = \frac{36}{49}\\\\x = \frac{-\frac{6}{7} +/- \sqrt{\frac{36}{49} } }{2 . -\frac{6}{7} } \\\\x = \frac{-\frac{6}{7} +/- \frac{6}{7} }{-\frac{12}{7} } \\\\x1 = \frac{-\frac{6}{7} + \frac{6}{7}}{-\frac{12}{7}} = 0\\\\x2 = \frac{-\frac{6}{7} - \frac{6}{7}}{-\frac{12}{7}} = \frac{\frac{12}{7}}{\frac{12}{7}} = 1$