Question

24. Resolva, em IR, as inequações.

a) ( 2x - 8 )(2 - x) > 0

b) (4x + 13)(3 - x)(2x - 1) < 0

c) x(3x – 4)(x + 2)(1 - x) < 0

Me Ajuda É Para Hoje,por favor​

Answer 1

Resposta:

(2x-8)(2-x) > 0(2x−8)(2−x)>0

4x-2 x^{2} -16+8x > 04x−2x

2

−16+8x>0

- 2x^{2}+12x-16 > 0−2x

2

+12x−16>0

a=-2::b=12::c=-16a=−2::b=12::c=−16

Por Báskara, temos que:

x= \frac{-b \frac{+}{} \sqrt{b ^{2}-4ac } }{2a}x=

2a

−b

+

b

2

−4ac

x= \frac{-12 \frac{+}{} \sqrt{12 ^{2}-4(-2)(-16) } }{2(-2)}x=

2(−2)

−12

+

12

2

−4(−2)(−16)

x= \frac{-12 \frac{+}{} \sqrt{144-128} }{-4}x=

−4

−12

+

144−128

x= \frac{-12 \frac{+}{} \sqrt{16} }{-4}x=

−4

−12

+

16

x= \frac{-12 \frac{+}{}4 }{-4}x=

−4

−12

+

4

x'= \frac{-12+4}{-4}::x'= \frac{-8}{-4}::x'=2x

′

=

−4

−12+4

::x

′

=

−4

−8

::x

′

=2

x''= \frac{-12-4}{-4}::x''= \frac{-16}{-4} ::x''=4x

′′

=

−4

−12−4

::x

′′

=

−4

−16

::x

′′

=4

espero ter ajudado bjss ❤❤❤