Question

已知A=2x的平方+3xy-y的平方,B-二分之一xy,C=八分之一x的三次方y的三次方-四分之一x的平方y的四次方。

求2AB的二次方-c
求求你们了，急a。。。。。。

Answer 1

2AB²-C
=2(2x²+3xy-y²)*(x²y²/4)-(x³y³/8-x²y^7/4)
=x^4y²+3x³y³/2-x²y^4/2-x³y³/8+x²y^7/4
=x^4y²+11x³y³/8-x²y^4/4

追问

x³y³/8-x²y^7/4？？

追答

哦，对不起，打错了
2AB²-C
=2(2x²+3xy-y²)*(x²y²/4)-(x³y³/8-x²y^4/4)
=x^4y²+3x³y³/2-x²y^4/2-x³y³/8+x²y^4/4
=x^4y²+11x³y³/8-x²y^4/4

Answer 2

A=2x^2+3xy-y^2
B=-xy/2
C=x^3y^3/8-x^2y^4/4

2AB^2-C
=2(2x^2+3xy-y^2)*(-xy/2)^2-(x^3y^3/8-x^2y^4/4)
=2(2x^2+3xy-y^2)*x^2y^2/4-(x^3y^3/8-x^2y^4/4)
=(4x^2+6xy-2y^2)*x^2y^2/4-(x^3y^3/8-x^2y^4/4)
=4x^2*x^2y^2/4+6xy*x^2y^2/4-2y^2*x^2y^2/4-(x^3y^3/8-x^2y^4/4)
=x^4y^2+3x^3y^3/2-x^2y^4/2-x^3y^3/8+x^2y^4/4
=x^4y^2+3x^3y^3/2-x^3y^3/8+x^2y^4/4-x^2y^4/2
=x^4y^2+(3/2-1/8)x^3y^3+(1/4-1/2)x^2y^4
=x^4y^2+11x^3y^3/8-x^2y^4/4

Answer 3

我不是他舅说的对

Answer 4

1425