?解:$?原式=(5-\frac {9}{4}-\frac {1}{4})xy-(\frac {9}{2}-\frac {1}{2})x^{3}y^{2}-x^{3}y?$ ?$=?\frac {5}{2}xy-4x^{3}y^{2}-x^{3}y$?
$??解:原式=(-1+4)(m-n)^{3}??+(2-1-1)(m-n)^{2} $ $=3(m-n)^{3}$
?$?$? 解:$?原式=\frac {1}{2}xy+2?$? ?$?把x=2,y=\frac {1}{4}代入原式=\frac {1}{2}×2×\frac {1}{4}+2=\frac {9}{4}?$?
?$?$? 解:由題,a+1=2,b-1=1,則a=1,b=2?
?$?原式=\frac {9}{2}a^{2}b?$? ?$?把a(bǔ)=1,b=2代入原式=\frac {9}{2}×1×2=9?$? 解:設(shè)第一步時(shí)，每堆牌的張數(shù)都是x(x≥2);第二步時(shí)，左邊有(x-2)張,中間有(x+2)張，右邊有x張;
?$第三步時(shí)，左邊有(x-2)張，中間有(x+3)張，右邊有(x-1)張;第四步開(kāi)始時(shí)，左邊有(x-2)張，$? ?$從中間拿走(x-2)張，則中間一堆牌的張數(shù)為(x+3)-(x-2)=x+3-x+2=5$? ?$所以中間一堆此時(shí)有5張牌$? 解:原式=1 化簡(jiǎn)后的結(jié)果中不含有x,所以x的取值與本題的計(jì)算結(jié)果無(wú)關(guān) ?解:$? 由題,a+1=b-2=0,則a=-1,b=2?$? ?$?原式=-a^{2}b^{2}+\frac {1}{2}ab+1?$? ?$?把a(bǔ)=-1,b=2代入原式=-1×4-1+1=-4?$?
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