Question

Пожалуйста помогите пжжпжпжп​

Answer 1

Ответ:

$\boxed{\ (a\pm b)^2=a^2\pm 2ab+b^2\ }\\\\1)\ \ x^2+2xy+y^2=(x+y)^2\\\\2)\ \ a^2+2a+1=(a+1)^2\\\\3)\ \ b^2-6b+9=b^2-2\cdot b\cdot 3+4^2=(b-3)^2\\\\4)\ \ c^2-10c+25=c^2-2\cdot c\cdot 5+5^2=(c-5)^2\\\\5)\ \ 4m^2+4m+1=(2m)^2+2\cdot (2m)\cdot 1+1^2=(2m+1)^2\\\\6)\ \ 16-8c+c^2=4^2-2\cdot 4\cdot c+c^2=(4-c)^2=(c-4)^2$