1/x

x/x+(1-x)/x-1/xx/x+(1-x)/x-1/xx/x+(1-x)/x-1/xx/[x+(1-x)/(x-1/x)]=x/[x+(1-x)/((x²-1)/x)]=x/[x+x(

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七年级下册政治复习提纲(山东人民出版社)第五单元是青春的脚步青春的气息别弄错了啊!格式：XXXXXXXXX1、XXXXXXXXXX.2、XXXXXXXXX.3、XXXXXXXXXXX.七年级下册政治复

(x+1)(x-1)(x-x+1)(x+x+1)(x+1)(x-1)(x-x+1)(x+x+1)(x+1)(x-1)(x-x+1)(x+x+1)(x+1)(x-1)(x-x+1)(x+x+1)=(x^

分解因式----（X*X-X)(X*X-X-2)+1分解因式----（X*X-X)(X*X-X-2)+1分解因式----（X*X-X)(X*X-X-2)+1(x-x)(x-x-2)+1=x(x-1)[

X(X-1)=XX(X-1)=XX(X-1)=XX(X-1)=Xx(x-1)-x=0x(x-1-1)=0x(x-2)=0x1=0,x2=2x∧2-2x=0(x-2)*x=0x=0x=2x(x-1+1)

1.1/x|x/(x+1)|1.1/x|x/(x+1)|1.1/x|x/(x+1)|1,x>1或x

x/x-1[1+(x-1/x)]x/x-1[1+(x-1/x)]x/x-1[1+(x-1/x)]（2x+1）/（x-1）

x-1/x÷(x-1/x)x-1/x÷(x-1/x)x-1/x÷(x-1/x)(x+1)/[(x2－1）／x〕x(x+1)/(x+1)(x-1)1/x÷(x-1/x)分子分母同×x得到1/(xx-1)

】x(x+1)+x^2(x-1)】x(x+1)+x^2(x-1)】x(x+1)+x^2(x-1)x(x+1)+x^2(x-1)=x^2+x+x^3-x^2=x+x^3

1/(1-x)+1/(1+x)+2/(1+x*x)+4/(1+x*x*x*x)+8/(1-x*x*x*x*x*x*x*x)怎么做1/(1-x)+1/(1+x)+2/(1+x*x)+4/(1+x*x*x

x>1/1-xx>1/1-xx>1/1-xx>1/1-xx(1-x)/(1-x)>1/(1-x)x(1-x)/(1-x)-1/(1-x)>0(x-x²-1)/(1-x)>0(x²-

x+1/x-1x+1/x-1x+1/x-1可用根轴法得-1到1

(X+1)/(x-1)(X+1)/(x-1)(X+1)/(x-1)(X+1)/(x-1)(X+1)/(x-1)(X+1)/(x-1)-1(X+1)/(x-1)-(x-1)/(x-1)(X+1-x+1)

(1+X)/(1-X)(1+X)/(1-X)(1+X)/(1-X)（字有点丑不要介意.）小于－1大于1x=±1-1小于-1或者大于1-1＜x＜1

|x+1|+|x-1||x+1|+|x-1||x+1|+|x-1|x＜-1时,-x-1-x+1≤1,x≥-1/2,此时无解-1≤x＜1时,x+1-x+1=2,此时无解x≥1时,x+1+x-1≤1,x≤

/X-3/-/X-1//X-3/-/X-1/什么是零点分段法?/X-3/-/X-1/零点分段法就是用函数的零点把数轴分成若干段,逐段讨论的解题方法.如本题中用x=1,x=3把数轴分成三段.(1).x原

化简|x+1/x|化简|x+1/x|化简|x+1/x|(1).0＜x＜1,x≠0,Ix+1/xI=x+1/x.(2).x＞1,x≠0,Ix+1/xI=x+1/x.(3).0＞x＞-1,x≠0,Ix+1

x-1/x-2x-1/x-2x-1/x-2当x>2时,x-1<2x-4,x>3,所以x>3当x<2时,x-1>2x-4,x<3,所以x<2综上所述,该不