ldcf.net
当前位置:首页 >> 用换元法解方程(x^2+1/x^2 )+5(x+1/x)%4+=0 >>

用换元法解方程(x^2+1/x^2 )+5(x+1/x)%4+=0

化为: y+1/y-2=0 y²+1-2y=0 (y-1)²=0 y=1

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

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...

let x= 2sinu dx =2cosu du ∫ x^2. (4-x^2)^(1/2) dx =∫ (2sinu)^2. (2cosu) (2cosudu) =16∫ (sinu)^2. (cosu)^2 du =4∫ (sin2u)^2 du =(1/8) ∫ (1-cos4u) du =(1/8)[u -(1/4)sin4u] + C =(1/8)[ arcsin(x/2) -(1/8)x(4-x^2)^(1/2) . [4- x^2(4-...

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=2 c-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=0 x=2±√3 当x+1/x=-1时 x+x+1=0 由于Δ=1-4*1=-3

设x2+1x=y.则原方程可化为y-2y+1=0.则y2-2+y=0.故答案为y2+y-2=0.

∵x?1x2=y,∴y+1y+2=0,整理得:y2+2y+1=0.故答案为:y2+2y+1=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.

网站首页 | 网站地图
All rights reserved Powered by www.ldcf.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com