计算﹕1-2x (-2分之3x)-(1-4分之x)

=3/3x-2/3x+x/(3x-2)=1/3x+x/(3x-2)=(3x-2+3x²)/3x(3x-2)=(3x²+3x-2)/(9x²-6x)

=x-2-x-三分之二=-3分之4

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

(x^2-1/9)÷(3x+1)=(x-1/3)(x+1/3)÷(3x+1)=(x-1/3)÷3=x/3-1/9

{[(x+2)/(x^2-2x)]-[(x-1)/(x-2)]}/[(x+3)/x]={[(x+2)/(x^2-2x)]-[(x^2-x)/(x^2-2x)]}*[x/(x+3)]=[(-x^2+2x

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

2/(x-3)=(1-x)/(3-x)-5,两边同乘以(x-3),得：2=-(1-x)-5(x-3),去括号,合并同类项,移项,得：4x=12,x=3.将x=3代入原方程检验,x-3=0,分母为0,分

x的平方-9分之1-x-3分之1除以x-1分之6+2x=1/(x^2-9)-1/(x-3)*(x-1)/2(3+x)=(2-(x-1))/2(x^2-9)=(3-x)/2(x^2-9)=-(x-3)/

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

思路：我们知道：1/3*2=(3-2)/3*2=3/3*2-2/3*2=1/2-1/3类比得：1/n*(n-1)=[n-(n-1)]/n*(n-1)=n/n*(n-1)-(n-1)/n*(n-1)=1

原式=x(x²-2x+3-3)×2/x²=x²(x-2)×2/x²=2x-4

x（x-1）分之1+(x-1)(x-2)分之1+(x-2)(x-3)分之1+...+（x-99）（x-100）分之1=1/(x-1)-1/x+1/(x-2)-1/(x-1)+...+1/(x-100)

原式=2x/(x^2-4)-1/(2-x)+3/(x+2)=2x/(x-2)(x+2)+1/(x-2)+3/(x+2)=[2x+(x+2)+3(x-2)]/(x-2)(x+2)=(6x-4)/(x-2

[x+1分之x]-[2x2-1分之2]=1再问：����̣�лл��

1-2x+(-2分之3x)-(1-4分之x)=1-2X-3\2X-1+1\4X=-2X-6\4X+1\4X=-2X-5\4X=-13\4X负4分之13X

先分别把第1、2项合并,第3、4项合并,最后加总,原式=[(5x+10)-(3x-3)]/[(x^2+x)-2]-[(5x+15)-(3x-6)]/[(x^2+x)-6]=(2x+13)/[(x^2+

原式=1/(x-3)-1/2(x+3)-x/(x-3)²=[2(x+3)(x-3)-(x-3)²-2x(x+3)]/[2(x+3)(x-3)²]=(2x²-18

原式=(x^2-8x+16)^3(x+1)^6/(x^2-2x+1)^3(x-4)^4(x+2)^2=(x-4)^6(x+1)^6/(x-1)^6(x-4)^4(x+2)^2=(x-4)^2(x+1)

裂项相消：1/[x(x+1)]+1/[(x+1)(x+2)]+1/[(x+2)(x+3)]+...+1/[(x+2011)(x+2012)]=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1

注:分子/分母(1/x-3)+(1/9-x^2)-(x-1/6-2x)通分:=2(x+3)/2(x+3)(x-3)+12/2(9-x^2)-(x-1)(3+x)/2(3-x)(3+x)=2x+6/2(