x-5分之x-4-x-6分之
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若原题是:5/2x+3=1/3+7x移项得5/2x-7x=1/3-3合并同类项得-9/2X=-8/3两边同乘以-2/9,得x=16/27.若原题是:(5x+3)/2=(1+7x)/3去分母得3(5x+
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
5分之3x-7<2分之x-23X/5-7
(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
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-
我来再问:再答:再答:就这样再答:谢谢采纳
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:【
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
(x-4)/(x-5)-(x-5)/(x-6)=(x-7)/(x-8)-(x-8)/(x-9)对方程两边各自通分:[(x-4)(x-6)-(x-5)(x-5)]/[(x-5)/(x-6)]=[(x-7
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
4分之5X-X+3=3分之2X-7-6分之X-25X/4-X+3=(2X–7)/3-(X-2)/615X-12X+36=4(2X–7)-2(X-2)3X+36=6X–243x=60X=20
(x+3)/7-(x+2)/5=(x+6)/6-(x+4)/4(1-2x)/35=-2x/2424-48x=-70x22x=-24x=-12/11
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)