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用换元法求3/x+1/(x-1)+4/(x-2)+4/(x-3)+1/(x-4)+3...

这道题要求计算能力很强 3/x+1/(x-1)+4/(x-2)+4/(x-3)+1/(x-4)+3/(x-5)=0 [3/x+3/(x-5)]+[1/(x-1)+1/(x-4)]+[4/(x-2)+4/(x-3)]=0 (6x-15)/(x^2-5x)+(2x-5)/(x^2-5x+4)+(8x-20)/(x^2-5x+6)=0 (2x-5)[3/(x^2-5x)+1/(x^2-5x+4)+4/(x^2-5x+6)]=0 所...

3/(x-2)+2/(x-3)=4/(x-1)+1/(x-4) (5x-6)/(x^2-5x+6)=(5x-17)/(x^2-5x+4) (5x-6)(x^2-5x+4)=(5x-17)(x^2-5x+6) 5x^3-31x^2-10x-24=5x^3-42x^2-75x-102 11x^2+65x+78=0 x=(-65±√793)/22

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∑为曲面z=3/4-(x2+y2) 求出偏导数为 zx=-2x,zy=-2y ∑在xoy面上的投影区域为 D:{(x,y)|x2+y2≤3/4} M=∫∫[∑]ρdS =∫∫[D]ρ√(1+zx2+zy2)·dxdy =∫∫[D]ρ√(1+4x2+4y2)·dxdy =∫[0~2π]dθ∫[0~√3/2]ρ√(1+4r2)·rdr =2π·ρ·∫[0~√3/2]√(1+4r2)·rdr =2π·ρ·1/8·∫[...

题目没错吗? 应该是:2x^4+x3-6x2-x-2 或者是2x^4-x3-6x2-x+2吧! 如果你确信题目没错,我想一下再给你答复!

y=【(x-1)(x+3)】【(x-2)(x+4)】+6 =(x^2+2x-3)(x^2+2x-8)+6 令x^2+2x-3=t t=(x+1)^2-4>=-4 y=(t)(t-5)+6 =t^2-5t+6 =(t-5/2)^2-1/4 t=(x+1)^2-4>=-4时 y>=-1/4 即值域为【-1/4,+∞)

1/(x-2)(x-3)-2/(x-1)(x-3)+1/(x-1)(x-2) =(x-1)/(x-2)(x-3)(x-1)-2(x-2)/(x-1)(x-2)(x-3)+(x-3)/(x-1)(x-2)(x-3) =【x-1-2x+4+x-3】/(x-1)(x-2)(x-3) =【0】/(x-1)(x-2)(x-3) =0

令y=x+1 ,则原式化成 (y-1)^2 + (1-1/y)^2 = 3 把平方展开y^2 - 2y + 1 + 1- 2/y + (1/y)^2 = 3 整理得 [y^2+2+(1/y)^2] - 2(y+1/y) = 3 (y+1/y)^2 - 2(y+1/y) - 3 = 0 再令z=y+1/y 带入,得 z^2 - 2z - 3 = 0 z=3 或者 z= -1 y+1/y= -1没有实...

解:令y=x/x-1,则原方程化为 y^2-3y+2=0 (y-1)(y-2)=0 得y1=1 y2=2 则有 x/x-1=1 此方程无解 x/x-1=2 解方程得x=2

解:设x=sinθ,则原式=∫(sinθ)^4dθ/(cosθ)^2。 而(sinθ)^4/(cosθ)^2=[1-(cosθ)^2]^2/(cosθ)^2=(secθ-cosθ)^2=(secθ)^2-2+(cosθ)^2=(secθ)^2-3/2+(1/2)cos2θ), ∴原式=∫[(secθ)^2-3/2+(1/2)cos2θ)]dθ=tanθ-3θ/2+(1/4)sin2θ+C, ∴原式=x/√(1-x^2)-...

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