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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...

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!

对于换元法来说其实这个提示已经比较充分了. 由P = x^4+x^3+x^2+x+1, 有x^5-1 = (x-1)(x^4+x^3+x^2+x+1) = (x-1)P. 于是(x^5+x^4+x^3+x^2+x+1)^2-x^5 = (x^5+P)^2-x^5 = x^10+2Px^5+P^2-x^5 = x^5·(x^5+2P-1)+P^2 = x^5·((x-1)P+2P)+P^2 = (x^5·...

令sinA=4/5,则cosA=3/5,那么原方程为(sinA)^x+(cosA)^x=1=(sinA)^2+(cosA)^2,可以得x=2 2011.4.12: 对于一般形式(a/c)^x+(b/c)^x=1 ,假定为m^x+n^x=1,由图像可证:m和n要么同时大于1,要么同时小于1。如果m和n同时大于1,则x是个负数;如果m...

令x^2-x=x(x-1)>0 x>1或 x2)(x^2-x)dx =(x^3/3-x^2/2)|(-1->0) -(x^3/3-x^2/2)|(0->1) +(x^3/3-x^2/2)|(1->2) =[(0-0)-(-1/3-1/2)]-[(1/3-1/2)-(0-0)]+[(8/3-4/2)-(1/3-1/2)] =5/6+1/6 +5/6 =11/6

你好 令arctanx=t,则x=tant,dx=sec²tdt ∫xe^arctanx/(1+x^2)^3/2dx =∫tante^t/(1+tan^2;t)^3/2*sec²tdt =∫tante^t/sec ³t sec ²tdt =∫tante^t/sectdt =∫sinte^tdt (1) =-∫e^tdcost =-coste^t+∫coste^tdt =-coste^t+sinte^...

(2)令t=tanx,x=arctant,dx=dt/(1+t^2) 原式=∫(0,√3) t/(1+t^2)dt =(1/2)*∫(0,√3) d(1+t^2)/(1+t^2) =(1/2)*ln|1+t^2||(0,√3) =ln2 (3)令t=x-π/2,x=t+π/2,dx=dt 原式=∫(-π/2,π/2) √[1+cos(2t+π)]dt =∫(-π/2,π/2) √(1-cos2t)dt =2*∫(0,π/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)-...

(x+1/2x-1)-(x-1/3)-1=0 解:(x+1/2x-1)-(x-1/3)-1=0 3/2x-1-x+1/3-1=0 3/2x-x=1-1/3+1 1/2x=5/3 x=10/3 希望对你有帮助。

[-1,1] 解析: //求y=x+√(1-x²)的值域 定义域:-1≤x≤1 设x=sint(-π≤x≤π),则: y =sint+cost =√2sin(t+π/4) ∵ -π≤t≤π ∴ -3π/4≤t+π/4≤5π/4 ∴ -√2/2≤sin(t+π/4)≤√2/2 ∴ -1≤√2sin(t+π/4)≤1 即,-1≤y≤1 值域是[-1,1]

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