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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...

∑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·∫[...

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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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,+∞)

(x-1)/(x+2)-3/(x-1)-1=0 (x-1)/(x+2)-[(x+2)-(x-1)]/(x-1)-1=0 (x-1)/(x+2)-(x+2)/(x-1)+1-1=0 (x-1)/(x+2)-(x+2)/(x-1)=0 设(x-1)/(x+2)=y,则y-1/y=0 即:y²-1=0,解得:y1=1 y2=-1 即:(x-1)/(x+2)=1,或者(x-1)/(x+2)=-1 解得:x=-1/2...

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

(1)(x+2) 2 +6(x+2)-91=O;设x+2=y,则原方程可变形为:y 2 +6y-91=0,解得:y 1 =7,y 2 =-13,当y 1 =7时,x+2=7,x 1 =5,当y 2 =-13时,x+2=-13,x 2 =-15;(2)x 2 -(1+2 3 )x-3+ 3 =0,[x-(3+ 3 )][x+(2- 3 )]=0,x-(3+ 3 ...

1. 设(√2+√3)x=t 1/t+t=4 t²-4t+1=0 t=2±√3 则 x=(2±√3)(√3-√2) x=2√3-2√2+3-√6 或 x=2√3-2√2-3+√6 2. 设3x+4=t 4/t+t/3=3 12+t²-9t=0 △=81-48=31 t=(9±√31)/2 代入 3x=(1±√31)/2 x=(1±√31)/6 3.设x-2y=u 2x+3y=v 方程组:2u/v+12=5u ...

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