1（3X^2+2XY-1/2X)-(2X^2-XY+x)2 (1/2XY+Y^2+1)+(X^2-1/2XY-2Y^2-1)3 -(X^2Y+3XY-4)+3(X^2Y-XY+2)4 -1/4(2K^3+4K^2-28)+1/2(K^3-2K^2+4K)

1（3X^2+2XY-1/2X)-(2X^2-XY+x)
=3X²-2X ²+2XY+XY-1/2X-X
=X²+3XY-3X/2
2 (1/2XY+Y^2+1)+(X^2-1/2XY-2Y^2-1)
=1/2XY-1/2XY+Y²-2Y²+X²+1-1
=X²-Y²
3 -(X^2Y+3XY-4)+3(X^2Y-XY+2)
=-X²Y+3X²Y+3XY-3XY-4+6
=2X²Y+2
4 -1/4(2K^3+4K^2-28)+1/2(K^3-2K^2+4K)
=1/2K³+1/2K³-K²-K²+7+2K
=K³-2K²+2K+7
这是我在静心思考后得出的结论，
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1（3X^2+2XY-1/2X)-(2X^2-XY+x)
解，得：
==（3X^2+2XY-1/2X)-(2X^2-XY+x)
==3x^2+2xy-1/2x-2x^2+xy-x
==3x^2-2x^2+2xy+xy-1/2x-x
==x^2+3xy-3/2x
2 (1/2XY+Y^2+1)+(X^2-1/2XY-2Y^2-1)
解，得：
=1/2xy-1/2xy+y^2+1+x^2-2y^2-1
=x^2-y^2
3 -(X^2Y+3XY-4)+3(X^2Y-XY+2)
解，得：
==-(x^2y+3xy-4)+3(x^2y-xy+2)
==-x^2y-3xy+4+3x^2y-3xy+6
==2x^2y-6xy+10
4 -1/4(2K^3+4K^2-28)+1/2(K^3-2K^2+4K)
解，得：
==-1/4(2k^3+4k^2-28)+1/2(k^3-2k^2+4k)
==-1/2k^3-k^2+7+1/2k^3-k^2+2k
==(1/2k^3-1/2k^3)-(k^2+k^2)+2k+7
==-2k^2+2k+7

(3x^2+2xy-1/2x)-(2x^2-xy+x)=3x^2+2xy-1/2x-2x^2+xy-x=3x^2-2x^2+2xy+xy-1/2x-x=x^2+3xy-3/2x（1/2xy+y^2+1)+(x^2-1/2xy-2y^2-1) =1/2xy-1/2xy+y^2+1+x^2-2y^2-1=x^2-y^2.-(x^2y+3xy-4)+3(x^2y-xy+2) =-x^2y-3xy+4+...

解什么，求什么？化简？