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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...

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) 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 ...

(X^2+1)/(x+1) + 3(X+1)/(x^2+1)=4 设 (X^2+1)/(x+1)=t t+3/t-4=0 t^2-4t+3=0 t=1,t=3 t=1==> (X^2+1)/(x+1)=1 ==>x^2-x==0,x=0,x=1 ,t=3 ==> (X^2+1)/(x+1)=3 ==>x^2-3x-2=0 x=(3+√17)/2 ,x=(3-√17)/2 解为 x=0,或x=1,或x=(3+√17)/2 ,或x=...

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)-...

该解答有错误.正确解答如下:设x+1=m,x-2=n,则原方程可以为:2m2+3mn-2n2=0即a=2,b=3n,c=-2n2∴m=?3n±9n2?4×2(?2n2)2×2=?3n±5n4,∴m1=12n,m2=-2n,∴x+1=12(x-2)或x+1=-2(x-2)∴x1=-4,x2=1.

设t=x^2+1 ∫x^3/(x^2+1) dx = ∫(x^2)/2(x^2+1) d(x^2+1) = ∫(t-1)/(2t) dt = ∫(1/2)dt - ∫(1/2t)dt = t/2 - (1/2)·lnt + C . = (x^2+1)/2 - (1/2)·ln(x^2+1) + C.

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