Question

(x+4)^2-(x-2)*(x+2) при x=0,125

Answer 1

$(x+4)^{2} -(x-2)$ × (x+2) при x=0,125
$(0,125+4)^{2} - (0,125-2)×(0,125+2)$
$0,125^{2} + 4^{2} - 0,125^{2} - 2^{2}$  ×  [/tex] 0,125^{2} + 2^{2} [/tex]
$4^{2} - 2^{2}$ × $0,125^{2} + 2^{2}$
16×0,015625=0,25