Question

(x-5)²=1 . Responde ai galera pfv cálculos

Answer 1

Fazendo :
$(x - 5) \times (x - 5) = {x}^{2} - 5x - 5x + 25 \\ {x}^{2} - 10x + 25$
Continuando:
${x}^{2} - 10x + 25 = 1 \\ {x}^{2} - 10x + 25 - 1 = 0 \\ {x}^{2} - 10x + 24 = 0$
Resolvendo a formula de Bhaskara temos:
$\frac{ - b + - \sqrt{ {b}^{2} - 4 \times ac} }{2a} \\ \frac{10 + - \sqrt{100 - 4 \times 1 \times 24} }{2} \\ \frac{10 + - \sqrt{4} }{2} \\ \frac{10 + 2}{2} = \frac{12}{2} = 6 \\ \frac{10 - 2}{2} = \frac{8}{2} = 4$
x' = 6 e x"= 4