Resolva as equações
Módulo inicial: Matemática A 10.º - Parte 1 - Pág. 38 Ex. 26
Enunciado
Resolva, em $\mathbb{R}$, as seguintes equações:
- $$\frac{{2\left( {x + 1} \right)}}{3} + 5\left( {x + 2} \right) = 8 – 3x$$
- $$3\left( {\frac{{x + 1}}{2} + \frac{{x – 1}}{3}} \right) = 5x – 2$$
- $$5 – \frac{{2\left( {x + 1} \right)}}{4} = \frac{{3x – 1}}{7}$$
- $$\frac{{x + 4}}{6} – \frac{{2\left( {x + 1} \right)}}{9} = \frac{{x – 2}}{6} + \frac{{11 – 2x}}{{18}}$$
- $$\left( {3x – \frac{2}{3}} \right)\left( {3x + \frac{2}{3}} \right) – 4 = {\left( {3x – 5} \right)^2} + \frac{5}{9}$$
- $$5{\left( {x – 2} \right)^2} – 500 = 0$$
Resolução
- Resolvendo a equação, vem:
$$\begin{array}{*{20}{l}} {\mathop {\frac{{2\left( {x + 1} \right)}}{3}}\limits_{(1)} + \mathop {5\left( {x + 2} \right)}\limits_{(3)} = \mathop 8\limits_{(3)} – \mathop {3x}\limits_{(3)} }& \Leftrightarrow &{2\left( {x + 1} \right) + 15\left( {x + 2} \right) = 24 – 9x} \\ {}& \Leftrightarrow &{2x + 2 + 15x + 30 = 24 – 9x} \\ {}& \Leftrightarrow &{26x = – 8} \\ {}& \Leftrightarrow &{x = – \frac{4}{{13}}} \end{array}$$ - Resolvendo a equação, vem:
$$\begin{array}{*{20}{l}} {3\left( {\frac{{x + 1}}{2} + \frac{{x – 1}}{3}} \right) = 5x – 2}& \Leftrightarrow &{\frac{{3x + 3}}{2} + x – 1 = 5x – 2} \\ {}& \Leftrightarrow &{3x + 3 + 2x – 2 = 10x – 4} \\ {}& \Leftrightarrow &{ – 5x = – 5} \\ {}& \Leftrightarrow &{x = 1} \end{array}$$ - Resolvendo a equação, vem:
$$\begin{array}{*{20}{l}} {\mathop 5\limits_{(28)} – \mathop {\frac{{2\left( {x + 1} \right)}}{4}}\limits_{(7)} = \mathop {\frac{{3x – 1}}{7}}\limits_{(4)} }& \Leftrightarrow &{140 – 14\left( {x + 1} \right) = 12x – 4} \\ {}& \Leftrightarrow &{140 – 14x – 14 = 12x – 4} \\ {}& \Leftrightarrow &{ – 26x = – 130} \\ {}& \Leftrightarrow &{x = 5} \end{array}$$ - Resolvendo a equação, vem:
$$\begin{array}{*{20}{l}} {\mathop {\frac{{x + 4}}{6}}\limits_{(3)} – \mathop {\frac{{2\left( {x + 1} \right)}}{9}}\limits_{(2)} = \mathop {\frac{{x – 2}}{6}}\limits_{(3)} + \mathop {\frac{{11 – 2x}}{{18}}}\limits_{(1)} }& \Leftrightarrow &{3\left( {x + 4} \right) – 4\left( {x + 1} \right) = 3\left( {x – 2} \right) + 11 – 2x} \\ {}& \Leftrightarrow &{3x + 12 – 4x – 4 = 3x – 6 + 11 – 2x} \\ {}& \Leftrightarrow &{ – 2x = – 3} \\ {}& \Leftrightarrow &{x = \frac{3}{2}} \end{array}$$ - Resolvendo a equação, vem:
$$\begin{array}{*{20}{l}} {\left( {3x – \frac{2}{3}} \right)\left( {3x + \frac{2}{3}} \right) – 4 = {{\left( {3x – 5} \right)}^2} + \frac{5}{9}}& \Leftrightarrow &{{{\left( {3x} \right)}^2} – {{\left( {\frac{2}{3}} \right)}^2} – 4 = 9{x^2} – 30x + 25 + \frac{5}{9}} \\ {}& \Leftrightarrow &{9{x^2} – \frac{4}{9} – 4 = 9{x^2} – 30x + 25 + \frac{5}{9}} \\ {}& \Leftrightarrow &{30x = 29 + 1} \\ {}& \Leftrightarrow &{x = 1} \end{array}$$RECORDE:
$$\left( {A + B} \right)\left( {A – B} \right) = {A^2} – B{}^2$$
$${\left( {A + B} \right)^2} = {A^2} + 2AB + {B^2}$$ - Resolvendo a equação, vem:
$$\begin{array}{*{20}{l}} {5{{\left( {x – 2} \right)}^2} – 500 = 0}& \Leftrightarrow &{{{\left( {x – 2} \right)}^2} – 100 = 0} \\ {}& \Leftrightarrow &{{x^2} – 4x + 4 – 100 = 0} \\ {}& \Leftrightarrow &{x = \frac{{4 \pm \sqrt {16 – 4 \times 1 \times \left( { – 96} \right)} }}{2}} \\ {}& \Leftrightarrow &{x = \frac{{4 \pm \sqrt {400} }}{2}} \\ {}& \Leftrightarrow &{x = \frac{{4 \pm 20}}{2}} \\ {}& \Leftrightarrow &{\begin{array}{*{20}{c}} {x = – 8}& \vee &{x = 12} \end{array}} \end{array}$$ALTERNATIVA:
$$\begin{array}{*{20}{l}} {5{{\left( {x – 2} \right)}^2} – 500 = 0}& \Leftrightarrow &{{{\left( {x – 2} \right)}^2} = \frac{{500}}{5}} \\ {}& \Leftrightarrow &{{{\left( {x – 2} \right)}^2} = 100} \\ {}& \Leftrightarrow &{\begin{array}{*{20}{c}} {x – 2 = – 10}& \vee &{x – 2 = + 10} \end{array}} \\ {}& \Leftrightarrow &{\begin{array}{*{20}{c}} {x = – 8}& \vee &{x = 12} \end{array}} \end{array}$$
