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2yx2 x4

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

x4-8y2(x2-2y2),=x4-8x2y2+16y4,=(x2-4y2)2,=[(x+2y)(x-2y)]2,=(x+2y)2(x-2y)2.

令x=tany ∫(x^2/(1+x^4))dx =∫(tany^2/(1+tany^4))*(1/(cosy)^2)dy =∫(siny)^2/((siny)^4+(cosy)^4) dy =∫(1/2)(1-cos2y)/(1-4(siny)^2(cosy)^2) dy =(1/2)∫(1-cos2y)/(1-(sin2y)^2) dy =(1/2)∫1/(1-(sin2y)^2) dy - (1/2)∫cos2y/(1-(sin2y)^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.

你好,可以根据极限公式解答, 极限是微积分中的基础概念,它指的是变量在一定的变化过程中,从总的来说逐渐稳定的这样一种变化趋势以及所趋向的值(极限值)。极限的概念最终由柯西和魏尔斯特拉斯等人严格阐述。在现代的数学分析教科书中,几乎...

x4-y4-x2y+3xy2-2x2y+2y4=x4+y4-3x2y+3xy2=(x4+y4)-3(x2y-xy2),当x4+y4=25,x2y-xy2=6时,原式=25-3×6=7.

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(1-2y+y²)-2x²(1-y²)+(y²+2y+1) =[x²(1-y)]²-2x²(1-y)(1+y)+(1+y)² =[x²(1-y)+(1+y)]² =(x²-x²y+1+y)²

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)

Z = x^4 + 3y^4 -2x^2y^3的二阶偏导数: Z"xy = -12xy^2 Z''yx = - 12xy^2 Z = x^4 + 3y^4 -2x^2y^3的一阶偏导数: Z'x = 4x^3 -4xy^3 Z'y = 12y^3 - 6x^2y^2 Z = x^4 + 3y^4 -2x^2y^3的二阶导数: Z"xx = 12x^2 - 4y^3 Z"yy = 36y^2 - 12x^2y 在数...

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