Question

(x+5) в квадрате - (х-2)(х+2)при х=0.125

Answer 1

(х+5)^2-(х-2)(х+2)=
=х^2+10х+25+х^2-4=
=2х^2+10х+21.... при х=0,125
=2*(0,125)^2+10*0,125+21=0,03125+1,25+21=
=22,28125