Resolve as inequações
Os números reais: Matematicamente Falando 9 - Pág. 105 Ex.33
Enunciado
Resolve as inequações:
- $-2x-3>3x-13$
- $3(x+2)<5(1+x)$
- $5(x+4)>2x$
- $12x-(x-1)\ge 7x$
- $5(1+3x)+\frac{1}{2}>5x$
- $\frac{1}{3}+\frac{1}{2}(x-1)>2x+1$
- $\frac{y+3}{6}\le 2-\frac{4-3y}{2}$
- $\frac{7x-3}{4}-\frac{9x-4}{8}>0$
- ${{(3+x)}^{2}}>{{x}^{2}}-1+7x$
Resolução
- Ora,
$$\begin{array}{*{35}{l}}
-2x-3>3x-13 & \Leftrightarrow & -2x-3x>-13+3 \\
{} & \Leftrightarrow & -5x>-10 \\
{} & \Leftrightarrow & \frac{-5x}{-5}<\frac{-10}{-5} \\
{} & \Leftrightarrow & x<2 \\
\end{array}$$
O conjunto solução é: \[S=\left] -\infty ,2 \right[\] - Ora,
$$\begin{array}{*{35}{l}}
3(x+2)<5(1+x) & \Leftrightarrow & 3x+6<5+5x \\
{} & \Leftrightarrow & -2x<-1 \\
{} & \Leftrightarrow & \frac{-2x}{-2}>\frac{-1}{-2} \\
{} & \Leftrightarrow & x>\frac{1}{2} \\
\end{array}$$
O conjunto solução é:\[S=\left] \frac{1}{2},+\infty \right[\] - Ora,
$$\begin{array}{*{35}{l}}
5(x+4)>2x & \Leftrightarrow & 5x+20>2x \\
{} & \Leftrightarrow & 3x>-20 \\
{} & \Leftrightarrow & \frac{3x}{3}>\frac{-20}{3} \\
{} & \Leftrightarrow & x>-\frac{20}{3} \\
\end{array}$$
O conjunto solução é:\[S=\left] -\frac{20}{3},+\infty \right[\] - Ora,
$$\begin{array}{*{35}{l}}
12x-(x-1)\ge 7x & \Leftrightarrow & 12x-x+1\ge 7x \\
{} & \Leftrightarrow & 4x\ge -1 \\
{} & \Leftrightarrow & \frac{4x}{4}\ge \frac{-1}{4} \\
{} & \Leftrightarrow & x\ge -\frac{1}{4} \\
\end{array}$$
O conjunto solução é:\[S=\left[ -\frac{1}{4},+\infty \right[\] - Ora,
$$\begin{array}{*{35}{l}}
5(1+3x)+\frac{1}{2}>5x & \Leftrightarrow & \underset{(2)}{\mathop{5}}\,+\underset{(2)}{\mathop{15x}}\,+\frac{1}{\underset{(1)}{\mathop{2}}\,}>\underset{(2)}{\mathop{5x}}\, \\
{} & \Leftrightarrow & 10+30x+1>10x \\
{} & \Leftrightarrow & 20x>-11 \\
{} & \Leftrightarrow & x>-\frac{11}{20} \\
\end{array}$$
O conjunto solução é:\[S=\left] -\frac{11}{20},+\infty \right[\] - Ora,
$$\begin{array}{*{35}{l}}
\frac{1}{3}+\frac{1}{2}(x-1)>2x+1 & \Leftrightarrow & \underset{(2)}{\mathop{\frac{1}{3}}}\,+\underset{(3)}{\mathop{\frac{x}{2}}}\,-\underset{(3)}{\mathop{\frac{1}{2}}}\,>\underset{(6)}{\mathop{2x}}\,+\underset{(6)}{\mathop{1}}\, \\
{} & \Leftrightarrow & 2+3x-3>12x+6 \\
{} & \Leftrightarrow & -9x>7 \\
{} & \Leftrightarrow & x<-\frac{7}{9} \\
\end{array}$$
O conjunto solução é:\[S=\left] -\infty ,-\frac{7}{9} \right[\] - Ora,
$$\begin{array}{*{35}{l}}
\frac{y+3}{\underset{(1)}{\mathop{6}}\,}\le \underset{(6)}{\mathop{2}}\,-\frac{4-3y}{\underset{(3)}{\mathop{2}}\,} & \Leftrightarrow & y+3\le 12-12+9y \\
{} & \Leftrightarrow & -8y\le -3 \\
{} & \Leftrightarrow & \frac{-8y}{-8}\ge \frac{-3}{-8} \\
{} & \Leftrightarrow & y\ge \frac{3}{8} \\
\end{array}$$
O conjunto solução é:\[S=\left[ \frac{3}{8},+\infty \right[\] - Ora,
$$\begin{array}{*{35}{l}}
\frac{7x-3}{\underset{(2)}{\mathop{4}}\,}-\frac{9x-4}{\underset{(1)}{\mathop{8}}\,}>\underset{(8)}{\mathop{0}}\, & \Leftrightarrow & 14x-6-9x+4>0 \\
{} & \Leftrightarrow & 5x>2 \\
{} & \Leftrightarrow & x>\frac{2}{5} \\
\end{array}$$
O conjunto solução é:\[S=\left] \frac{2}{5},+\infty \right[\] - Ora,
$$\begin{array}{*{35}{l}}
{{(3+x)}^{2}}>{{x}^{2}}-1+7x & \Leftrightarrow & 9+6x+{{x}^{2}}>{{x}^{2}}-1+7x \\
{} & \Leftrightarrow & -x>-10 \\
{} & \Leftrightarrow & x<10 \\
\end{array}$$
O conjunto solução é:\[S=\left] -\infty ,10 \right[\]
