Question

若x=2分之1，y=1，求x(x的2次方+xy+y的2次方)-y(x的2次方+xy+y的2次方)+3xy(y-x)

若x=2分之1，y=1，求x(x的2次方+xy+y的2次方)-y(x的2次方+xy+y的2次方)+3xy(y-x)的值

Answer 1

x(x的2次方+xy+y的2次方)-y(x的2次方+xy+y的2次方)+3xy(y-x)
=x³+x²y+y³-x²y-xy²-y³+3xy²-3x²y
=x³+2xy²-3x²y
=x(x²-3xy+2y²)
=x(x-y)(x-2y)
因为
x=1/2,y=1
所以
原式=1/2×（1/2-1）×（1/2-2）=3/8

Answer 2

x(x^2+xy+y^2)-y(x^2+xy+y^2)+3xy(y-x)
=x^3+x^2y+xy^2-x^2y-xy^2-y^3+3xy^2-3x^2y
=x^3-y^3+3xy^2-3x^2y
=(x-y)(x^2+xy+y^2)+3xy(y-x)
=(x-y)(x^2+xy+y^2-3xy)
=(x-y)(x-y)^2
=(x-y)^3
=(1/2-1)^3
=(-1/2)^3=-1/8

Answer 3

-八分之一

Answer 4

x^2+y^2=7 xy=-2
(y^2-x^2)^2=(y^2+x^2)^2-4x^2y^2=49-16=33 y^2-x^2=±√33
5x^2-3xy-11xy-7x^2+2y^2
=2y^2-2x^2-14xy=2(y^2-x^2)+28=28±2√33