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用换元法解方程(x^2+1/x^2 )+5(x+1/x)%4+=0

因为y=x+1x,所以y2=(x+1x)2,整理得x2+1x2+2=y2,即:x2+1x2=y2-2.所以原方程可化为y2+y-2=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...

化为: y+1/y-2=0 y²+1-2y=0 (y-1)²=0 y=1

由原方程,得x2+(1x)2+5(x+1x)-66=0,∴x2+2+(1x)2+5(x+1x)-66-2=0,即(x+1x)2+5(x+1x)-68=0,∵设x+1x=t,∴原方程可化为t2+5t-68=0.故答案是:t2+5t-68=0.

设:x/(x+1)为m m²-2m-8=0 (m-4)(m+2)=0 m1=4 m2=-2 x/(x+1)=4 x=4x+4 3x=-4 x=-4/3 x/(x+1)=-2 x=-2x-2 3x=-2 x=-2/3 x1=-4/3 x2=-2/3

∵x^-1/x-x/x^-1+2=0 y=x^-1/x ∴x^-1/x=y x/x^-1=1÷(x^-1/x)=1/y 则原式=y-1/y+2=0 y^2+2y-1 D 望采纳哦

因为x2+1x2=(x+1x)2-2,所以原方程可整理为y2-2-2y-1=0,进一步整理得:y2-2y-3=0.

把x2+1x=y代入原方程,得y-4×1y+3=0,方程两边同乘以y整理得:y2+3y-4=0.故选A.

y-1/y+2=o 方程两边各乘以y,则有 y²+2y-1=0 求解就好了

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.

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