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用换元法解方程(x^2+1/x^2 )+5(x+1/x)%4+=0

1、题目应是 2/(x-1)+4/(x^2-1)/4=0 2/(x-1)+4/(x-1)(x+1)=0 2/(x-1)=t x-1=2/t x+1=2/t +2 (x+1)/2=(1+t)/t 2/(x+1)=t/(1+t) 则方程变为 t+t*t/(1+t)=0 t(1+t)+t^2=0 2t^2+t=0 t(2t+1)=0 t=0(舍去) t=-1/2 2/(x-1)=-1/2 1-x=4 x=-3 经验证x...

化为: y+1/y-2=0 y²+1-2y=0 (y-1)²=0 y=1

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...

(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...

解:(1)令: ,则原方程变为: 即 解得: 分别代入 得 ,解得 得 解得 ;原方程的解是: , ;(2)令: 则原方程变为: 解得 由于 因此 不合题意,舍去将 代入 ,得 解得 原方程的解为 。

令它等于一个新数(x),反解,代入,注意新元定义域

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.

换元法解方程⑴ 2(x^2+1)/(x+1)+6(x+1)/(x^2+1)=7 令(x^2+1)/(x+1)=t 2t+6/t=7 2t^2-7t+6=0 (2t-3)(t-2)=0 t=3/2或t=2 (1)t=3/2 (x^2+1)/(x+1)=3/2 2(x^2+1)=3(x+1) 2x^2+2=3x+3 2x^2-3x-1=0 x=(3±√3^2+4*2*1)/2*2=(3±√17)/4...

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