Question

5x^y-[2x^2y-3(xy^2-2x^2y+2xy^2)],其中x,y满足|x-1|+(y+2)^2=0

先化简，再求值
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Answer 1

5x^y-[2x^2y-3(xy^2-2x^2y+2xy^2)]
=5x^y-[2x^2y-3xy^2+6x^2y-6xy^2]
=5x^y-2x^2y+3xy^2-6x^2y+6xy^2
=5x^y-8x^2y+9xy^2

|x-1|+(y+2)^2=0
所以 x-1=0 y+2=0

x=1 y=-2

代入上式得 5*1^-2-8*(-2)+9*(-2)^2
=5+16+36
=21+36
=57

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Answer 2

∵|x-1|+(y+2)^2=0
∴x-1=0 x=1 y+2=0 y=-2
5x^y-[2x^2y-3(xy^2-2x^2y+2xy^2)]
=5x^y-[2x²y－3(3xy²-2x²y)]
＝5x^y-[2x²y－9xy²＋6x²y]

＝5x^y-[8x²y－9xy²]

＝5x^y-8x²y＋9xy²

＝5×1^﹙－2﹚－8×1²×﹙-2﹚＋9×1×﹙-2﹚²

＝5+16+36

＝57