Question

: 1.(3а+4)^2=   (2х-b)=   (b+3)(b-3)=    (5y-2x)(5y+2x)=

Answer 1

$(3a+4)^2=(3a)^2+2*(3a)*4+4^2=9a^2+24a+16$
$(2x-b)=2x-b$
$(2x-b)^2=(2x)^2-2*(2x)*b+b^2=4x^2-4bx+b^2$
$(b+3)(b-3)=b^2-3^2=b^2-9$
$(5y-2x)(5y+2x)=(5y)^2-(2x)^2=25y^2-4x^2$