解一道分式方程! X\/(X-1)-(X-1)\/(X-2)=(X-3)\/(X-4)-(X-4)\/(X-5)
X\/(X-1)-(X-1)\/(X-2)=(X-3)\/(X-4)-(X-4)\/(X-5)(X-1+1)\/(X-1)-(X-2+1)\/(X-2)=(X-4+1)\/(X-4)-(X-5+1)\/(X-5)1+1\/(X-1)-[1+1\/(X-2)]=1+1\/(X-4)-[1+1\/(X-5)]1\/(X-1)-1\/(X-2)=1\/(X-4)-1\/(X-5)...
你会解方程x\/x-1-x-1\/x-2=x-3\/x-4-x-4\/x-5
先通分再求解 (x^2-2x-x^2+2x-1)\/(x-1)\/(x-2)=(x^2-8x+15-x^2+8x-16)\/(x-4)\/(x-5)1\/(x-1)\/(x-2)=1\/(x-4)\/(x-5)x^2-9x+20=x^2-3x+2 6x=18 x=3
解方程:(x-1)\/(x-2)-(x-3)\/(x-4)=(x-2)\/(x-3)(x-4)\/(x-5)
1\/(x-2)-1\/(x-4)=1\/(x-3)-1\/(x-5)-2\/[(x-2)(x-4)]=-2\/[(x-3)(x-5)](x-2)(x-4)=(x-3)(x-5)x^2-6x+8=x^2-8x+15 2x=7 x=3.5 经检验,它是原方程的根
解方程:(x-1)\/(x-2)-(x-3)\/(x-4)=(x-2)\/(x-3)(x-4)\/(x-5)
整理:-2\/(x-2)(x-4)=-2\/(x-3)(x-5)得到:(x-2)(x-4)=(x-3)(x-5)多项式乘多项式: x2-6x+8=x2-8x+15 解得:x=3.5 最后分式方程要检验
数学解方程 x\/x-1-3\/(x-1)(x+2)=1
数学解方程 x\/x-1-3\/(x-1)(x+2)=1 我来答 1个回答 #热议# 富含维C的水果为何不能做熟吃?南瓜灯上降落 推荐于2021-01-17 · TA获得超过394个赞 知道小有建树答主 回答量:884 采纳率:0% 帮助的人:185万 我也去答题访问个人页 关注 展开全部 更多追问追答 追问 文字 两边...
解方程(x-1\/x-2) -(x-3\/x-4)=(x-2\/x-3)-(x-4\/x-5)
很简单吧。例如(x-1)\/(x-2)=1- 1\/(x-2)将这四项都这么变形,原方程变为 1\/(x-4) -1\/(x-2)=1\/(x-5)-1\/(x-3)两边通分,得1\/(x-2)(x-4)=1\/(x-5)(x-3)所以(x-2)(x-4)=(x-3)(x-5)拆开得-6x+8=-8x+15 解得x=3.5 ...
解分式方程:x\/(x-1)-3\/(1-x)=3
x\/(x-1)-3\/(1-x)=3 x\/(x-1)+3\/(x-1)=3 两边乘x-1 x+3=3(x-1)x+3=3x-3 x=3 经检验,x=3是方程的解
解方程(x+1)\/(x-2)+(x-1)\/(x-3)=(x-3)\/(x-4)+(x+3)\/(x-1)
先通分,不用乘开来。得到 ((x+1)(x-3)+(x-1)(x-2))\/(x-2)(x-3)=((x-3)(x-1)+(x+3)(x-4))\/(x-1)(x-4)斜对角分别相乘。得到新式移项 得到(x-3)(x-4)(-x+5)=(X-1)(X-2)(-X-5)再约掉-x-5 后面不用我说了吧 这种问题不要怕,看到...
求函数y=(x-1)*(x-2)^2* (x-3)^3*(x-4)^4的拐点,求详细解题方法。我使...
首先,说说图是怎么画的,这种幂相乘连续函数,一笔就可以画完,在数轴上找到0点,有1,2,3,4,四个点,取X趋向无穷大时,显然y是无穷大,所以由x=4的右方开始画,x=1,2,3,4时,Y=0,所以用光滑曲线向点(4,0)画,不穿过(因为x-4是4次幂,领域内符号相同,且对称)如图示,同理,...
求高人解方程式 要过程(x\/x-1)-1=3\/(x-1)(x+2) 急急急!!!
x\/(x-1)-1=3\/[(x-1)(x+2)]方程两边同时乘以(x-1)(x+2)得 x(x+2)-(x-1)(x+2)=3 x^2+2x-x^2-x+2=3 x=1 经检验,x=1为增根 则此方程无解