"(x+1)^2-4(x+1)+4" x-1=3的算木平方根

问题描述:

"(x+1)^2-4(x+1)+4" x-1=3的算木平方根

x^2y^2(x-4y)(x+4y)-x^2y^2(x-2y)^2 =x^2y^2(x^2-16y^2-x^2+4xy-4y^2)
=x^2y^2(4xy-20y^2)
=4x^2y^3(x-5y)
当x=-2,y=1/2时,原式=4×(-2)^2×(1/2)^3×(-2-5×1/2)=-9