[(x+y)^5-2(x+y)^6+4(-x-y)^3+(-x-y)^2]÷1/2(x+y)^2（4x^2-9y^2)/(2x+3y)=已知a=-1,求代数式[5a^4-(-3a^6)^2+12a^5b÷3ab]÷（-2a^2）^2的值

Question

[(x+y)^5-2(x+y)^6+4(-x-y)^3+(-x-y)^2]÷1/2(x+y)^2
（4x^2-9y^2)/(2x+3y)=
已知a=-1,求代数式[5a^4-(-3a^6)^2+12a^5b÷3ab]÷（-2a^2）^2的值

Answer 1

[(x+y)^5-2(x+y)^6+4(-x-y)^3+(-x-y)^2]÷1/2(x+y)^2
=[(x+y)^5-2(x+y)^6-4(x+y)^3+(x+y)^2]÷1/2(x+y)^2
=(x+y)^3-4(x+y)^4-8(x+y)+2
（4x^2-9y^2)/(2x+3y)
=(2x+3y)(2x-3y)/(2x+3y)
=2x-3y
[5a^4-(-3a^6)^2+12a^5b÷3ab]÷（-2a^2）^2
=[5a^4-9a^12+4a^4]÷4a^4
=5/4-9/4a^8+1
=9/4-9/4a^8
=9/4-9/4x(-1)^8
=9/4-9/4
=0
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