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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

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...

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

x4+y4+(x+y)4-2,=(x2+y2)2-2x2y2+(x2+2xy+y2)2-2,=(x2+y2)2-2x2y2+(x2+y2)2+4xy(x2+y2)+4x2y2-2,=2(x2+y2)2+2x2y2+4xy(x2+y2)-2,=2[(x2+y2)2+x2y2+2xy(x2+y2)-1],=2[(x2+xy+y2)2-1],=2(x2+xy+y2-1)(x2+xy+y2+1)...

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) ...

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^4-2y^4-2x³y+xy³ =x^4-2x³y+xy³-2y^4……(用加法交换律) =x³(x-2y)+y³(x-2y)……………(提取公因式) =(x-2y)(x³+y³)…………………(提取公因式) =(x-2y)(x+y)(x²-xy+y²)………(立方和公式)

解:当x≥2y时,z=x-y,画出区域图平移直线x-y=0,当过点A(-2,-1)时,直线y=x-z的截距最大,此时z最小最小值为z=-2-(-1)=-1当x<2y时,z=x4+y2,画出区域图平移直线y=0,当过点A(-2,-1)时,直线y=-x2+2z的截距最小,此时z最小最小值为z=-...

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

∵x+y=-1,∴x4+5x3y+x2y+8x2y2+xy2+5xy3+y4,=(x4+2x2y2+y4)+5xy(x2+y2)+xy(x+y)+6x2y2,=(x2+y2)2+5xy[(x+y)2-2xy]+xy(x+y)+6x2y2,=[(x+y)2-2xy]2+5xy(1-2xy)-xy+6x2y2,=(1-2xy)2+5xy-10x2y2-xy+6x2y2,=1-4xy+4x2y2+5xy-10...

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