[12(x+y)³+3(x+y)(x-y)²]÷6(x+y)

问题描述:

[12(x+y)³+3(x+y)(x-y)²]÷6(x+y)

(x+y)0时,分子分母同时除以(x+y),再拆去里面的小括号 则 原式=(12x^2+12y^2+3x^2+3y^2+18xy)/6=2.5(x^2+y^2+3xy)

[12(x+y)³+3(x+y)(x-y)²]÷6(x+y)
=3(x+y)[4(x+y)²+3(x-y)²]÷6(x+y)
=[4(x+y)²+3(x-y)²]÷2
=[4x²+8xy+4y²+3x²-6xy+3y²]÷2
=(7x²+2xy+7y²)/2

[12(x+y)³+3(x+y)(x-y)²]÷6(x+y)
=2(x+y)²+(x-y)²/2