已知x-3分之x-2=x-3分之m
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(1)y=x/x(x-1)*(x-1)²/(x-1)(x+1)-2/(x+1)=1/(x+1)-2/(x+1)=-1/(x+1)=1/3∴x+1=-3,x=-4即x=-4时,y=1/3(2)
A/(x-3)+1/(x+4)=[A(x+4)+(x-3)]/(x-3)(x+4)=(Ax+4A+x-3)/(x-3)(x+4)=[(A+1)x+(4A-3)]/(x-3)(x+4)∵(2x+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/(x+10
解题思路:先解方程求出x,再化简另一分式并把x值代入计算即可解题过程:解:经检验是原方程的解。
右边通分=[A(x-2)-B]/(x-2)²=[Ax+(-2A-B)]/(x-2)²=(x+3)/(x-2)²所以Ax+(-2A-B)=x+3A=1-2A-B=3所以A=
2-(X-1)/3=(1-X)/2+3-X(7-X)/3=(7-3X)/214-2X=21-9X7X=7X=1把X=1代入方程得4-K/3=0K=12
原式=[(1-1分之x+5)-(1-1分之x+4))+(1-1分之x+3)-(1-1分之x+2)]*[(x+3)(x+5)]\(x^2+7x+13)=[(1分之x+2)-(1分之x+3)+(1分之x+
已知分式4x^2-1分之6x^2+x-2的值为0,则x^-1=∵分式4x^2-1分之6x^2+x-2的值为0∴6x²+x-2=04x²-1≠0(3x+2)(2x-1)=0∴3x+2
由于A/(X+6)+B/(X-3)=(AX-3A+BX+6B)/(X+6)(X-3)=(2X+1)/(x+6)(x-3)所以,A+B=2,6B-3A=1,解得,A=11/9B=7/9
[(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+
x/(x^2+x+1)=-1/2即-2x=x^2+x+1x^2+3x+1=0(x-1)/(2x^2+x^3-1)/(x^2-x+1)=(x-1)/(2x^2+x^3-1)(x^2-x+1)=(x-1)
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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
x-1=1-3^(1/2)
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