解方程:x² x分之7 x²-x分之3=x²-1分之6
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/16 03:06:05
X²-3X分之5X-7=X-1分之2+X-2分之3X²-3X分之5X-7=(X-1)(X-2)分之[2(X-2)+3(X-1)]X²-3X分之5X-7=(X²-
x平方+x分之7+x平方-x分之3=x平方-1分之61分之6应该就是6吧?x²+7/x+x²-3/x=x²-6x²+x²-x²+7/x-3/
x+7分之1x=56两边同乘以7,得7x+x=56×78x=56×7x=7×7x=49x-4分之x=364分之3x=364分之x=12x=48
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)/(
5x/(x²+x-6)+(2x-5)/(x²-x-12)=(7x-10)/(x²-6x+8)5x/(x-2)(x+3)+(2x-5)/(x+3)(x-4)=(7x-10)
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+
(x^2+x)分之7-(x^2-1)分之6=(x-x^2)分之37/(x^2+x)-6/(x^2-1)=3/(x-x^2)7/[x(x+1)]-6/[x(x-1)]=-3/[x(x-1)]分母最小公倍
这个容易阿,将x=5带入整个方程,得出结果m等于3.满意请采纳
x+x/7=24/497x+x=24/78x=24/7x=24/7/8x=3/7
(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
(x-3分之x)+(x^2-3x分之x+6)=x分之x-3x/(x-3)+(x+6)/x(x-3)=(x-3)/x两边同乘以x(x-3)x²+x+6=x²-6x+97x=3x=3/
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
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-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
方程两边同乘以x﹙x+1﹚﹙x-1﹚消去分母得:7﹙x-1﹚+3﹙x+1﹚=6x展开整理化简得:x=1经检验x=1方程的增根,∴原方程无解.
化简3x/5+2x/5-3x/7=4/3x-3x/7=4/34x/7=4/3x=7/3
我知道再答:
两边乘以(x+2)(x-2)得x(x-2)-(x-2)²=kx²-2x-x²+4x-4=k2x=k+4∵无解∴x=-2或2当x=-2时k=-8当x=2时k=0
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