当x^2-4x+1=0时，求x^2/(x-1)-[1+1/(x^2-x)]的值
答案:2 悬赏:70 手机版
解决时间 2021-02-14 13:50
- 提问者网友：情歌越听越心酸
- 2021-02-14 07:30
当x^2-4x+1=0时，求x^2/(x-1)-[1+1/(x^2-x)]的值
最佳答案
- 五星知识达人网友：毛毛
- 2021-02-14 08:25
x^2-4x+1=0
x^2=4x-1
x^2/(x-1)-[1+1/(x^2-x)]
=(4x-1)/(x-1)-(x^2-x+1)/(x^2-x)
=(4x-1)/(x-1)-(4x-1-x+1)/(4x-1-x)
=(4x-1)/(x-1)-3x/(3x-1)
=[(4x-1)(3x-1)-3x(x-1)]/(x-1)(3x-1)
=(9x^2-4x+1)/(3x^2-4x+1)
=[9(4x-1)-4x+1]/[3(4x-1)-4x+1)
=(32x-8)/(8x-2)
=8(4x-1)/2(4x-1)
=4
x^2=4x-1
x^2/(x-1)-[1+1/(x^2-x)]
=(4x-1)/(x-1)-(x^2-x+1)/(x^2-x)
=(4x-1)/(x-1)-(4x-1-x+1)/(4x-1-x)
=(4x-1)/(x-1)-3x/(3x-1)
=[(4x-1)(3x-1)-3x(x-1)]/(x-1)(3x-1)
=(9x^2-4x+1)/(3x^2-4x+1)
=[9(4x-1)-4x+1]/[3(4x-1)-4x+1)
=(32x-8)/(8x-2)
=8(4x-1)/2(4x-1)
=4
全部回答
- 1楼网友：西风乍起
- 2021-02-14 09:30
解：x²-4x+1=0
x=（-b±√b²-4ac）/2a=（4±√14-4/）2=（4±√10）/2
∴x=（4+√10）/2或x=（4-√10）/2
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