limsinxln(1+x)

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

(x+1)(x-1)(x-x+1)(x+x+1)(x+1)(x-1)(x-x+1)(x+x+1)=(x^2-1)(x+1-x)(x+1+x)=(x^2-1)[(x+1)^2-x^2]=x^2(x+1)^2-x^4-(x+1)^2+x^2=x

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

X(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)=0x=0x平方-x-x=0x平方-2x=0提取x（x-2）=0x=0x

1.1/x|x/(x+1)|1,x>1或x

x/x-1[1+(x-1/x)]（2x+1）/（x-1）

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)。然后分母就可以用平方差公式解开（x+1)(x-1)。这么讲解可以吗？

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

X+X*(1-X)+X*(1-x)²+X*(1-x)³+.+X*(1-x)ⁿ=想问下有没有什么数学公示的?这个利用等比数列求和公式①当X=1时,X+X*(1-X)+X*(1-x)²+X*(1-x)&

x*x+2x+x*x+10x+x*x+25+2x+x*x+1+1+16+x*x+4x*x+9-12x+16x*x+4-16x=30x*x-7x-15求x,可以只要答案.如图,全是正方形,一个边长为1,一个边长为4,求总面积,你也可以用你的方

x-1)(X-2)(x-3)...(x-50)+x(x-2)(X-3)...(X-50)+...+x(x-1)(x-2)...(x-49)等于多少?等于(x-1)(x-2)(x-3)(x-4)...(x-50)的导数

x>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-1)>0x²-x+1=(x-1/2)

x+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)-1(X+1)/(x-1)-(x-1)/(x-1)(X+1-x+1)/(x-1)2/(x-1)将右边的1移到左边，则变为（x+1）/(x-1)-(x-1

(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≤1/2,此时无解综上,|x+1|+|x-1|≤1的解集是空集x

/X-3/-/X-1/什么是零点分段法?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原式成为(3-x)-(1-x)2无解.(2).1≤x原式成为(3-x)-(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/xI=-x-1/x.(4).x＜-1,x≠0,Ix+1/xI=-x-1/x

x-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,该不等式的解为x>3或x<2两边同时-2则3-x/x-2＜0

{2X+1}+{X}这种题得分情况讨论①当x|2x+1|+|x|2x+1+x3xx