Question

resolva as equações
a)x/2-5x+6-<0
b)-2x/2+x+1<0

Answer 1

$A)~~ \dfrac{x}{2} -5x+6\ \textless \ 0\to~~mmc=2\\\\ x-10x+12\ \textless \ 0\to\\\\ -9x\ \textless \ -12\to~(-1)\\\\ 9x\ \textgreater \ 12\to\\\\x\ \textgreater \ \dfrac{12}{9} \to\\\\x\ \textgreater \ \dfrac{4}{3}\\\\\\ \large\boxed{\boxed{S = \left\{x ~\epsilon~ \mathbb{R} \ | \ x\ \ \textgreater \ \ \frac{4}{3} \right\}}}$





$B)~~ \dfrac{-2x}{2+x} +1\ \textless \ 0\to~~\\\\\\ \dfrac{-2x}{2+x} \ \textless \ -1\to~~multiplique~~cruzado:\\\\\\ -2x\ \textless \ -2-x\to\\\\ -2x+x\ \textless \ -2\to\\\\ -x\ \textless \ -2\to\\\\ x\ \textgreater \ 2\\\\\\ \large\boxed{\boxed{S = \left\{x ~\epsilon~ \mathbb{R} \ |~x\ \textgreater \ 2 \right\}}}$

Answer 2

As inequações não estão bem definidas
Vou fazer com uma das posibilidades

a)
               $\frac{x}{2-5x} +6 \ \textless \ 0 \\ \\ \frac{x}{2-5x} \ \textless \ -6 \\ \\ x\ \textless \ -6(2-5x) \\ \\ x\ \textless \ -12+30x \\ \\ x-30x\ \textless \ -12 \\ \\ -29x\ \textless \ -12 \\ \\ 29x\ \textgreater \ 12 \\ \\ x\ \textgreater \ \frac{12}{29}$

                                         S = {x∈R| $x\ \textgreater \ \frac{12}{29}$ }

b)
               $\frac{-2x}{2+x}+1\ \textless \ 0 \\ \\ \frac{-2x}{2+x} \ \textless \ -1 \\ \\ -2x\ \textless \ -1(2+x) \\ \\ -2x\ \textless \ -2-x \\ \\ -2x+x\ \textless \ -2 \\ \\ -x\ \textless \ -2 \\ \\ x\ \textgreater \ 2$

                                         S = { x∈R| x > 2 }