计算(1-2x)²-x²＝0

(x-1/x)平方-3x+3/x+2=0(x-1/x)^2-3(x-1/x)+2=0(x-1/x-2)(x-1/x-1)=0(x-1/x-2)=0或(x-1/x-1)=0(1)x-1/x-2=0x^2

f(x)=1*x^0+2*x^1+.+n*x^(n-1)+.xf(x)=x+...+(n-1)x^(n-1)+nx^n+...f(x)-xf(x)=1+x+x^2+x^3+...x^(n-1)-nx^

X=1/2.结果=2005

=x²+3x-(x²-3x+2)=x²+3x-x²+3x-2=6x-2

3/(x-2)-2/(x-1)=3(x-1)/(x-2)(x-1)-2(x-2)/(x-2)(x-1)=[3(x-1)-2(x-2)]/(x-2)(x-1)=(3x-3-2x+4)/(x-2)(x-1

=6x²-10x-3x+5=6x²-13x+5如果您认可我的答案,请点击下面的“选为满意回答”按钮,【供参考,\(^o^)/~】

0*∞的不定型,先化成比值,然后洛必达ln(sinx)=-------------------x^(-1/2)无穷/无穷洛必达(1/sinx)*cosx=-------------------(-1/

(x^2-1)=（x+1）*（x-1）(x^2+x)=x（x+1）通分x-(2x-1)/x=（(x^2-2x+1）/x=（x-1)^2/x带回写成分式后发现约掉一个x,一个（x+1）,一个（x-1）化

(x^2+5x+4)(x^2+5x+6)=(x^2+5x)^2+10(x^2+5x)+24=x^4+10x^3+35x^2+50x+24

longfun(intx,intn)//这个函数计算x的n次方{inti;longq=1;if(n==0){return1;}if(n==1){returnx;}for(i=1;i

(x-1)(x^2+x+1)-x(x^2-2x+4)=x^3-1-x^3+2x^2-4x=2x^4-4x-1如果本题有什么不明白可以追问,另外发并点击我的头像向我求助,请谅解,

原式=lim(x→0)(1+x)^(3/(2x)+2)=lim(x→0)(1+x)^(3/(2x))=lim(x→0)[(1+x)^(1/x)]^(3/2)=e^(3/2)

3x1x-1用十字相乘法（3X+1)(X-1)=0X1=-1/3,X2=1

1/X(X+1)+1/(X+1)(X+2)+.+1/(X+99)(X+100)=[1/x-1/(x+1)]+[1/(x+1)-1/(x+2)]+.+[1/(x+99)-1/(x+100)]=1/x-1

1/x(x+1)+1/(x+1)(x+2)+.+1/(x+2010)(x+2011)=1/x-1/(x+1)+1/(x+1)-1/(x+2)+.+1/(x+2010)-1/(x+2011)=1/x-1

（x+1)(x-2)(x-3)(x-6)==（x+1)(x-6)(x-2)(x-3)=(x^2-5x-6)(x^2-5x+6)=(x^2-5x)^2-36=x^4-10x^3+25x^2-36

1/x[x+1]化成差值形式1/x-1/(x+1)

(2^x+x+1)²=2^（x+1）+x²+1+x*2^(x+1)+2^(x+1)+2x+1

(X-1)(X-2)(X-3)(X-4)=(X-1)(X-4)(X-2)(X-3)=（x^2-5x+4)(x^2-5x+6)=(x^2-5x)^2+10(x^2-5x)+24=x^4-10x^3+25

【参考答案】原式=[(x^2+x)/(x^2-1]-[x^2/(x^2-2x+1)]=[(x^2+x)(x-1)/(x^2-1)(x-1)]-[x^2(x+1)/(x^2-2x+1)(x+1)]=[(