Question

Resolva as inequações, com U = Q.
URGENTE!!

Answer 1

Resposta:

não faço ideia mas preciso de pontos mil desculpas

Answer 2

Resposta:

Solução:

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf -\:\dfrac{1}{3} \cdot \left ( \dfrac{3}{2} - 2x \right ) > \dfrac{1}{2} \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf -\:\dfrac{3}{6} + \dfrac{2x}{3} > \dfrac{1}{2} \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf -\: \dfrac{3}{6} + \dfrac{4x}{6} > \dfrac{3}{6} \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf -\;3 + 4x > 3 \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf 4x > 3 + 3 \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf 4x > 6 \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf x > \dfrac{6}{4} \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf x > \dfrac{3}{2} \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf U = \bigg\{x\in\mathbb{Q} \mid x> \dfrac{3}{2} \bigg\} \end{array}\right$

f)

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf \dfrac{x+ 1}{3} -\: \dfrac{x+ 1}{5} \geq \dfrac{x+1}{10} \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf \dfrac{10\cdot(x+ 1)}{30} -\: \dfrac{6 \cdot (x+ 1)}{30} \geq \dfrac{3\cdot (x+1)}{30} \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf 10\cdot (x+1)-6\cdot (x + 1) \geq 3\cdot (x +1) \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf 10x +10 -6x -6 \geq 3x + 3 \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf 10x -6x - 3x\geq 3 +6 - 10 \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf 10x -9x \geq 9 - 10 \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf x \geq -\; 1 \end{array}\right$

$\large \sf \displaystyle \left\begin{array}{l l l l l } \sf U = \big\{x\in\mathbb{Q} \mid x \geq -\:1 \big\} \end{array}\right$

Explicação passo-a-passo:

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