2分之1(6-x)2x=8
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X+1分之X+2+X+7分之X+8=X+5分之X+6+X+3分之X+4等式两边各2个x常数和都为10所以都约去得1/x+7/x=5/x+3/x为恒等式所以x为不等于零的任意实数再问:������ڶ�Щ
理解为如下等式:2/x+6/(x+1)=18/[x(x+1)]2(x+1)+6x=182x+2+6x=188x=16x=2
请加括号再问:没有括号再答:是这样吗?x/(x-2)+x/(x-7)-9=x/(x-1)+1+x/(x-6)-8再问:是:x/(x-2)+(x-9)/(x-7)=(x+1)/(x-1)+(x-8)/(
(1)2x/(x+1)=12x=x+1x=1(2)x/(x-2)-1/(x²-4)=1x(x+2)-1=x²-4x²+2x-1=x²-42x=-3x=-3/2(
x+1分之x+2减去x+3分之x+4=x+5分之x+6减去X+7分之x+8方程两边分别通分,得[(x+2)(x+3)-(x+1)(x+4)]/[(x+1)(x+3)]=[(x+6)(x+7)-(x+8
x+2分之x+1+x+9分之x+8=x+3分之x+2+x+8分之x+7即x+2分之1+x+9分之1=x+3分之1+x+8分之1即x+2分之1-x+3分之1=+x+8分之1-x+9分之1(x+2)(x+
化简等式得:1-1/(X+6)+1-1/(x+3)=1-1/(x+2)+1-1/(x+7),整理得1/(x+6)-1/(x+7)=1/(x+2)-1/(x+3);1/[(x+6)(x+7)]=1/[(
x平方-6x+8=x²-2×3x+9-9+8=(x-3)²-1=(x-3-1)(x-3+1)=(x-4)(x-2)x平方+x-6=(x-2)(x+3)12+x-x平方=(4-x)(
(5-x分之1-x)-(1-x分之5-x)=x^2-6x+5分之8(x-1)/(x-5)-(x-5)/(x-1)=8/(x-5)(x-1)(x-1)²-(x-5)²=8x²
(x-3)分之x-(x^2-3x)分之(x+6)+x分之1等于几?=x²/x(x-3)-(x+6)/x(x-3)+(x-3)/x(x-3)=(x²-x-6+x-3)/x(x-3)=
左右都有X^2,故把它们都约去,剩下X分之6=X+1两边都乘以x得6=x平方+x移项x平方+x-6=0因式分解(x-3)(x+2)=0解得x=3或x=-2
等号前后同时分别计算减法得:(x+1)(x+3)分之2=(x+5)(x+7)分之2所以:(x+1)(x+3)=(x+5)(x+7)x^2+4x+3=x^2+12x+35x=-4
再问:再答:
(x+1)/(x+2)-1/(x+7)=(x+2)/(x+3)-1/(X+6)(X+2-1)/(X+2)-1/(X+7)=(X+3-1)/(X+3)-1/(X+6)1-1/(X+2)-1/(X+7)=
2x+4/3x-1+x-3/6-2x=07/3x=9/6x=9/143x-1/6=4x-3/8x=3/8-1/6x=5/24
x+2分之x+1-x+3分之x+2=x+6分之x+5-x+7分之x+61-(x+2)分之1-[1-(x+3)分之1]=1-(x+6)分之1-【1-(x+7)分之1】从而(x+3)分之1-(x+2)分之
本人一次只回答一个算了吧或者你分开提问x/(x+2)=x/(x-1)两边乘(x+2)(x-1)x²-x=x²+2x3x=0x=0经检验,x=0是方程的解
X=4原式=(X-1)/(X-2)-(X-2)/(X-3)=(X-4)/(X-5)-(X-5)/(X-6)左边通分=[X^2-4X+3-X^2+4X-4]/[(X-2)(X-3)]=-1/[(X-2)