已知A=2x^2+3xy-y^2,B=(-1/2)xy,C=(1/8)x^3y^3-(1/4)x^

已知A=2x^2+3xy-y^2,B=(-1/2)xy,C=(1/8)x^3y^3-(1/4)x^2y^4,求2AB^2-C的值

解:
2AB²-C
=2(2x²+3xy-y²)(-1/2xy)²-(1/8x³y³-1/4x²y^4)
=2(2x²+3xy-y²)(x²y²)/4-(1/8x³y³-1/4x²y^4)
=x^4y^2+3/2*x^3y^3-1/2*x^2y^4-(1/8x³y³-1/4x²y^4)
=x^4y^2+11/8*x^3y^3-1/4*x^2y^4
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第1个回答  2014-03-27
A=1
B=7/4
C=0

A-{B+[C-(A+B)]}=2