先化简,再求值:x²-1 x² 2x 1除以x-1 x² x,其中x=2016-搜答案-智慧作业帮

原式=(1+x²)/(1+x)(1-x)÷(2x-x+x²)/(1-x)=(1+x²)/(1+x)(1-x)×(1-x)/x(x+1)=(1+x²)/x(1+x

原式=(x−1)(x+1)−x(x−2)x(x+1)×(x+1)2x(2x−1)，=2x−1x(x+1)×(x+1)2x(2x−1)=x+1x2，∵x2-x-1=0，∴x2=x+1，将x2=x+1代入

原式=x²-x+2x²+2x-6x²+15x+2x-5=-3x²+18x-5当x=2时原式=-12+36-6=19

分母[x-1-(2x-1)/(x+1)]=(x²-1-2x+1))/(x+1)=(x²-2x)/(x+1)[(x²-2x)/(x²-1)]/[x-1-(2x-1

2（x-3)(x+2)+(3+x)(3-x)-(x-1)的平方=2x²-2x-12+9-x²-x²+2x-1=-4如果本题有什么不明白可以追问,

（1x2+4x+4+1x2+2x）÷x+1x+2=[1(x+2)2+1x(x+2)]×x+2x+1=2(x+1)x(x+2)2×x+2x+1=2x(x+2)，当x=3时，原式=23×(3+2)=215

x+1/x除以（x-1+x²/2x）=(x+1)/x÷(2x^2-1-x^2)/2x=2(x+1)/(x^2-1)=2(x+1)/(x+1)(x-1)=2/(x-1)已知x=√2+1x-1=

(3x/x-1-x/x+1)×x^2-1/x=[3x(x+1)-x(x-1)]/[(x-1)(x+1)]*(x^2-1)/x=(2x^2+4x)/x=2x+4因为x=(根号2)-2,所以原式=2*(√

(3x/x+1-x/x-1)+x-2/x的平方-1=(3x(x-1)-x(x+1))/(x+1)(x-1)+x-2/x^2-1=(3x^2-3x-x^2-x+x-2)/(x^2-1)=(2x^2-3x

x^2+(4x-3x^2)-(5x-6x^2)-2=x^2+4x-3x^2-5x+6x^2-2=4x^2-x-2=1-1/2-2=-3/2

原式=(x-3)/(x+1)(x-1)×(x+1)²/(x-3)=(x+1)/(x-1)当x=-6/5时原式=(-1/5)/(-11/5)=1/11

原式=2x²+x-3-4x²-4x-1+4x²+4x=2x²+x-4=8-2-4=2

原式=1x+2−(x−2)2x(x−1)÷(x+1)(x−1)−3x−1=1x+2−(x−2)2x(x−1)•x−1(x+2)(x−2)=1x+2−x−2x(x+2)=2x2+2x，由x2+2x-4=

(1)(x-1)(x+4)-(x-3)(x+2)=x²+3x-4-(x²-x-6)=x²+3x-4-x²+x+6=4x+2∴当x=1/3时,原式=4×(1/3)

原式=2x+1÷4x(x+1)=2x+1×x(x+1)4=x2，当x=-2时，原式=-22=-1．

原式=(x+2)(x+3)x+3•x−3(x+2)(x−2)−1=x−3x−2−x−2x−2=−1x−2；当x=3时，原式=−13−2=2+3．

您好：(X-1/X平方-1)-(X/X+1)=（x-1）/(x+1)(x-1)-x/(x+1)=1/(x+1)-x/(x+1)=(1-x)/(x+1)=(1-2)/(2+1)=-1/3如果本题有什么不

能不能先用（）把每个分数括起来?

原式=2(x^2-1)-2x^2+x=2x^2-2-2x^2+x=x-2代入得原式=-4希望我的回答可以令您满意