[4x+(1/2)]的平方

[4x+(1/2)]^2=[4x+(1/2)]*[4x+(1/2)]=4x*4x+4x*(1/2)++(1/2)*4x++(1/2)*(1/2)=16x^2+4x+1/4[(1/4)m-2n]^2=[(1/4)m-2n]*[(1/4)m-2n]=(1/4)m*(1/4)m-(1/4)m*2n-2n*(1/4)m+2n*2n=(1/16)m^2 -mn+4n^2