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用换元法求3/x+1/(x-1)+4/(x-2)+4/(x-3)+1/(x-4)+3/(x-5)=0

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 所以2x-5可以等于0 所以x

设y=x-3/x+1变为y+1/y=2.等号两边同乘y解得y=x-3/x+1=1但x-3/x+1=1无解,所以这个题无解

你好!用换元法解方程2(x+1)+3(x+1)(x-2)-2(x-2)=0令x+1=a,x-2=b原式为2a+3ab-2b=0(a+2b)(2a-b)=0将x+1和x-2代入(x+1+2x-4)(2x+2-x+2)=0(3x-3)(x+4)=0得x=-4或x=-1检验:x=1时原式为 8-6-2=0 x=-4时原式为 18+3(-3)(-6)-72=0打字不易,采纳哦!

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-10211x^2+65x+78=0x=(-65±√793)/22

(x-1)(x-2)(x-3)(x-4)-3=(x-5x+4)(x-5x+6)-3令x-5x=t=(t+4)(t+6)-3=t+10t+24-3=t+10t+21=(t+3)(t+7)=(x-5x+3)(x-5x+7)中间令x-5x+5=t也可以=(t-1)(t+1)-3=t-

x+1/x-3(x+1/x)=(x+2+1/x)-2-3(x+1/x)=(x+1/x)-2-3(x+1/x)=2设x+1/x为c则有c-2-3c=2c-3c-4=0所以c=4或-1即x+1/x=4或-1当x+1/x=4时x-4x+1=0x=2±√3当x+1/x=-1时x+x+1=0由于Δ=1-4*1=-3<0所以不成立所以原方程的解为x=2±√3

(X+1)(X+2)(X+3)(X+4)-24 =(X+1)(X+4)(X+2)(X+3) -24 =(x+5x+4)(x+5x+6)-24 令x+5x=t,可得 (t+4)(t+6)-24 =t+10t+24-24 =t+10t =t(t+10) 即 原式 =(x+5x)(x+5x+10) =x(x+5)(x+5x+10)

第一题:(x-1)(x-4)(x-2)(x-3)=(x^2-5x+4)(x^2-5x+6) 设y=x^2-5x,则原式=(y+4)(y+6)-24=y^2+10y=y(y+10) 把y带入,则原式=(x^2-5x)(x^2-5x+10)=x(x-5)(x^2-5x+10) 第二题:原式=x^3-2x^2+X^2-4=x^2(x-2)+(x-2)(x+2)=(x-2)(x^2+x+2)

1.原式=(x^2+5x+4)(x^2+5x+6)-24-----------【两两相乘】 =(x^2+5x+5)^2-1-24-------------------【平方差公式】 =(x^2+5x+5-5)(x^2+5x+5+5)---------【因式平方差公式】 =(x+5)(x^2+5x+10)x2.原式=(x-6)(x-9)(x+4)(x+7)+350-----------【十字

x^2+1/x^2 +x +1/x =4 (x+1/x)^2 +(x+1/x)-6=0 (x+1/x+3)(x+1/x-2)=0 => x+1/x=-3 or x+1/x=2 x^2+3x+1=0 or x^2-2x+1=0 x=(-3+√5)/2 or (-3-√5)/2 or 1

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