x²+4x+4分之x²-x-6

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

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

x^2-x-6分解因式得(x+2)(x-3)x^2-4分解因式得(x-2)(x+2)故(x^2-x-6分之x^2-4+x-3分之x+2)=(x-2)/(x-3)+(x+2)/(x-3)=2x/(x-3

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)

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

解分式方程输入很辛苦.经检验X=7/2是原方程的解.

希望能帮上忙,第一步x-7/x-9+x-3/x-5=x-4/x-6+x-6/x-8第二步(x-9)+2/x-9+(x-5)+2/x-5=(x-6)+2/x-6+(x-8)+2/x-8第三步1+2/x-

题目表达不清楚

[(x-2)分之3x-(x+2)分之x]乘x分之(x²-4)=[(x-2)分之3x]乘x分之(x²-4)-[(x+2)分之x]乘x分之(x²-4)=3(x+2)-(x-2

由(x-4)/(x-5)-(x-5)/(x-6)=(x-7)/(x-8)-(x-8)/(x-9)得：（X-4）/（X-5）-（X-7）/（X-8）=（X-5）/（X-6）-(X-8）/（X-9）即：【

4-4x+x^2分之2x-6除以（x+3)乘以3-x分之x^2+x-6=2(x-3)/(x-2)²×1/(x+3)×(x+3)(x-2)/(3-x)=-2/(x-2)

因为(x-4)/(x-5)=[(x-5)+1]/(x-5)=1+1/(x-5)……所以原方程可化为1+1/(x-5)-1-1/(x-6)=1+1/(x-8)-1-1/(x-9)1/(x-5)-1/(x

(x+5)/(x+4)+(x+6)/(x+5)=(x+4)/(x+3)+(x+7)/(x+6)1+1/(x+4)+1+1/(x+5)=1+1/(x+3)+1+1/(x+6)1/(x+4)+1/(x+5

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

解原式=3(2-x)/3(x-4)(x+2)+(x+4)(x-4)/3(x+2)(x-4)=(6-3x)/3(x-4)(x+2)+(x²-16)/3(x+2)(x-4)=(6-3x+x

每个分式减去1(x-4-x+5)/(x-5)+(x-8-x+9)/(x-9)=(x-7-x+8)/(x-8)+(x-5-x+6)/(x-6)1/(x-5)+1/(x-9)=1/(x-8)+1/(x-6

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+2分之1+x^2-4分之x+6-x^-4x+4分之2x+1=x-2/x^2-4+x+6/x^2-4-(2x+1/x^-4x+4)=2x+4/x^2-4--(2x+1/x^-4x+4)=2/x--2

（x-2分之x减x+2分之x）=(x-2)(x+2)分之x(x+2)-x(x-2)乘4x分之2-x=(x-2)(x+2)分之4x乘4x分之2-x=-x-2