Question

Me ajudem por favor!

Answer 1

$\boxed{\boxed{Ola\´ \: \: Lucas} }$

$a) \: \: \left| \begin{array}{ccc} x -2 & 6 \\ 3 & 5 \end{array} \right| = 2$
$\Delta = 2 \\ \Leftrightarrow 5(x -2) -3 \cdot 6 = 2 \\ \Leftrightarrow 5x - 10 - 18 = 2 \\ \Leftrightarrow 5x -28 = 2 \\ \Leftrightarrow 5x = 30 \\ \Leftrightarrow x = \frac{30}{5}$
$\Leftrightarrow \boxed{\boxed{x= 6} }} \end{array}\qquad\checkmark$



$b) \: \: \left| \begin{array}{ccc} 2 & 3 & -2 \\ 0 & 1 & x \\ 2 & x & -3 \end{array} \right| = 2$

$\left| \begin{array}{ccc} 2 & 3 & -2 \\ 0 & 1 & x \\ 2 & x & -3 \end{array} \right| \: \left | \begin{array}{ccc} 2 & 3 \\ 0 & 1 \\ 2 & x \end{array} \right| = 2 \\ \Delta = 2 \\ \Leftrightarrow ( - 6 + 6x + 0) - ( - 4 + 2 {x}^{2} + 0) = 2 \\ \Leftrightarrow - 6 + 6x + 4 - 2 {x}^{2} = 2 \\ \Leftrightarrow - 2 {x}^{2} + 6x - 2 - 2 = 0 \\ \Leftrightarrow - 2 {x}^{2} + 6x - 4 = 0 \\ \Rightarrow - x {x}^{2} + 3x - 2 = 0$
• Os coeficientes são:
$\begin{cases} a = -1 \\ b = 3 \\ c = -2 \end{cases} \\ \\ \Delta = b^2 - 4ac \\ \Leftrightarrow \Delta = 3^2 - 4.(-1).(-2) \\ \Leftrightarrow \Delta = 9 -8 \\ \Leftrightarrow \Delta = 1 \\ \\ x = \frac{-b \pm \sqrt{\Delta} }{2a} \\ \\ \Leftrightarrow x = \frac{-3 \pm \sqrt{1} }{2.(-1)} \\ \begin{cases} x_1 = \frac{-3+1}{-2} \\ x_2 = \frac{-3 -1}{-2} \end{cases} \\ \Leftrightarrow \\ \begin{cases} x_1 = \frac{-2}{-2} \\ x_2 = \frac{-4}{-2} \end{cases} \\ \Leftrightarrow \\ \begin{cases} x_1 = 1 \\ x_2 = 2 \end{cases}$

Solução: $x \in \{ 1; 2 \}$

$\textbf{Bons estudos}$ !