先化简再求值[1/x+1-(x+3/x^2-1)*(x^2+1/x^2+4x+3)]/2/x+1其中x=3的平方根-1
[1/册携(x+1)-(x+3)/(x^2-1)*(x^2-1)/(x^2+4x+3)]/[2/(x+1)]
=[1/睁做(x+1)-(x+3)/[(x^2-1)](x^2-1)/[(x+3)(x+1)]]/[2/(x+1)]
=[1-(x^2-1)/(x-1)]/州早伏2
=1/2-(x+1)/2
=(1-√3)/2
![先化简再求值[1/x+1-(x+3/x^2-1)*(x^2+1/x^2+4x+3)]/2/x+1其中](/img/547601708.jpg)
先化简再求值[1/x+1-(x+3/x^2-1)*(x^2+1/x^2+4x+3)]/2/x+1其中
先化简再求值[1/x+1-(x+3/x^2-1)*(x^2+1/x^2+4x+3)]/2/x+1其中x=3的平方根-1
[1/册携(x+1)-(x+3)/(x^2-1)*(x^2-1)/(x^2+4x+3)]/[2/(x+1)]
=[1/睁做(x+1)-(x+3)/[(x^2-1)](x^2-1)/[(x+3)(x+1)]]/[2/(x+1)]
=[1-(x^2-1)/(x-1)]/州早伏2
=1/2-(x+1)/2
=(1-√3)/2