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2yx2 x4

这题目条件不够,y和x之间的关系没有,y和x^2的阶数可能一样可能不一样 总的来说,解题思路是这样子的,分子分母都除以分子,这样子得到 lim(x-0,y-0).1/(x^2/y+y/x^2) 这样子的话,y和x^2趋于0时,极小数阶数一致的话,那答案就是1/2

-x^4+2x^2y^2-y^4 =-(x^2-y^2)^2 =-(x+y)^2(x-y)^2

不存在,令y=kx^2,将其代入原式, 原式=lim(x→0)x^2(kx^2)/x^4+(kx^2)^2 =lim(x→0)kx^4/x^4+k^2x^4 =lim(x→0)k/1+k^2 =k/(1+k^2) 所以极限不存在

-x4y2+4x3y3-4x2y4=-x2y2(x2-4xy+4y2)=-x2y2(x-2y)2.原式=-22×(?12)2=-1.

令y=x^2,得2x^2y/(x^4+y^2)=1, 令y=-x^2,得2x^2y/(x^4+y^2)=-1. 由极限的唯一性知,所求极限不存在。

x^4(x-2y)+x²(2y-x) =x^4(x-2y)-x²(x-2y) =x²(x-2y)(x²-1) =x²(x-2y)(x-1)(x+1)

x^4+(x+y)^4+y^4 =(x^2+y^2)^2-2x^2y^2+(x+y)^4 =[(x+y)^2-2xy]^2-2x^2y^2+(x+y)^4 =[(x+y)^2]^2-4xy(x+y)^2+4x^2y^2-2x^2y^2+(x+y)^4 =2(x+y)^4-4xy(x+y)^2+2x^2y^2 =2[(x+y)^4-2xy(x+y)^2+(xy)^2] =2[(x+y)^2-xy]^2 =2(x^2+xy+y^2)^2

∵x4+y4+2x2y2-x2-y2-12=0,∴(x2+y2)2-(x2+y2)-12=0,即(x2+y2+3)(x2+y2-4)=0,∴x2+y2=-3,或x2+y2=4,∵x2+y2≥0,∴x2+y2=4.

x4-y4-x2y+3xy2-2x2y+2y4 =x4+y4-3x2y+3xy2 =(x4+y4)-3(x2y-xy2) =25-3x6 =25-18 =7

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