Question

已知x-y=1,求x²-y²+x-3y的值

再补一个问题吧……[利用因式分解求x+3y=125(x≠y)时,(x²+2xy-3y²)÷(x-y)的值]

Answer 1

x²-y²+x-3y=(x+y)(x-y)+x-3y
=x+y+x-3y
=2x-2y
=2(x-y)
=2

(x²+2xy-3y²)÷(x-y)
=(x²+2xy+y²-y²-3y²)÷(x-y)
=[(x+y)²-4y²]÷(x-y)
=[(x+y-2y)(x+y+2y)]÷(x-y)
=x+3y
=125

Answer 2

x²-y²+x-3y
=（x-y)(x+y)+x-y-2y
=x+y+1-2y
=x-y+1
=1+1
=2

(x²+2xy-3y²)÷(x-y)
=(x²+2xy+y²-y²-3y²)÷(x-y)
=[(x+y)²-4y²]÷(x-y)
=[(x+y-2y)(x+y+2y)]÷(x-y)
=(x-y)(x+3y)÷(x-y)
=x+3y
=125

Answer 3

X=1+Y
则 原式=(1+Y)^2-Y^2+(1+Y)-3Y
=1+2Y+Y^2-Y^2+1+Y-3Y
=2

X^2+2XY-3Y^2=(X-Y)×(X+3Y)
所以原式=(X-Y)×(X+3Y)÷(X-Y)
=125