通分一，x平方-x+3分之6x+1与-x平方+x-3分之4x-3 二（x-1）（x-2）分之3x与（x-1）（2-x）分之3+x_百度知道
通分一，x平方-x+3分之6x+1与-x平方+x-3分之4x-3 二（x-1）（x-2）分之3x与（x-1）（2-x）分之3+x
一、(6x+1)/(x^2-x+3)(4x-3)/(-x^2+x-3)=(3-4x)/(x^2-x+3)二、3x/[(x-1)(x-2)]=3x/(x^2-3x+2)(3+x)&#攻丹缔诽郫赌惦涩定绩47;[(x-1)(2-x)]=-(x+3)/(x^2-3x+2)
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通分一，x平方-x+3分之6x+1=(6x+1)/(x²-x+3)与-x平方+x-3分之4x-3 =-(4x-3)/(x²-x+3攻丹缔诽郫赌惦涩定绩)二（x-1）（x-2）分之3x=3x/(x-1)(x-2)（x-1）（2-x）分之3+x=-(x+3)/(x-1)(x-2)
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解一题：-x²+x-3=-(x²-x+3)x²-x+3与-x²+x-3的最简公分母是(x²-x+3)x平方-x+3分之6x+1=(6x+1)/(x²-x+3)-x平方+x-3分之4x-3=-(4x-3)/(x²-x+3)解二题：(x-1)(2-x)=-(x-1)(x-2)(x-1)(x-2)与(x-1)(2-x)的最简公分母是(x-1)(x-2)(x-1)(x-2)分之3x=3x/[(x-1)(x-2)](x-1)(2-x)分之3+x=-(3+x)/[(x-1)(x-2)]
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出门在外也不愁对于试题：“先化简，再求值：x-3x2-1-11-x，其中x=2．”小亮写出了如下解答过程：∵x-3x2-1-11-x=x-3(x-1)(x+1)-1x-1①=x-3(x-1)(x+1)-x+1(x-1)(x+1)②=x-3-（x+1）=2x-2，③∴当x=2时，-数学试题及答案
繁体字网旗下考试题库之栏目欢迎您！
1、试题题目：对于试题：“先化简，再求值：x-3x2-1-11-x，其中x=2．”小亮写出了如..
发布人：繁体字网（） 发布时间： 7:30:00
对于试题：“先化简，再求值：x-3x2-1-11-x，其中x=2．”小亮写出了如下解答过程：∵x-3x2-1-11-x=x-3(x-1)(x+1)-1x-1①=x-3(x-1)(x+1)-x+1(x-1)(x+1)②=x-3-（x+1）=2x-2，③∴当x=2时，原式=2×2-2=2．&&&&&&&&&&& &④（1）小亮的解答在哪一步开始出现错误：______（直接填序号）；（2）从②到③是否正确：______；若不正确，错误的原因是______；（3）请你写出正确的解答过程．
&&试题来源：不详
&&试题题型：解答题
&&试题难度：中档
&&适用学段：初中
&&考察重点：分式的加减乘除混合运算及分式的化简
2、试题答案：该试题的参考答案和解析内容如下：
（1）①；（2）不正确；把分母去掉了；（3）正确的应是：x-3x2-1-11-x=x-3(x-1)(x+1)+x+1(x-1)(x+1)=2x+1；当x=2时，原式=23．
3、扩展分析：该试题重点查考的考点详细输入如下：
&&&&经过对同学们试题原文答题和答案批改分析后，可以看出该题目“对于试题：“先化简，再求值：x-3x2-1-11-x，其中x=2．”小亮写出了如..”的主要目的是检查您对于考点“初中分式的加减乘除混合运算及分式的化简”相关知识的理解。有关该知识点的概要说明可查看：“初中分式的加减乘除混合运算及分式的化简”。
4、其他试题：看看身边同学们查询过的数学试题：
1、2、3、4、5、6、7、8、9、10、11、12、13、14、15、16、17、18、19、20、21、22、23、24、25、26、27、28、29、30、31、32、33、34、35、36、37、38、39、40、41、42、43、44、45、46、47、48、49、50、51、52、高一数学：已知集合A=｛x l x（x²-3x+2）=0｝，B=｛x l (x-1)(x-2)(x-3)=0｝，C=｛x l mx=1｝定_百度知道
高一数学：已知集合A=｛x l x（x²-3x+2）=0｝，B=｛x l (x-1)(x-2)(x-3)=0｝，C=｛x l mx=1｝定
已知集合A=｛x l x（x²-3x+2）=0｝B=｛x l (x-1)(x-2)(x-3)=0｝C=｛x l mx=1｝.定义集合M与N新运算：M※N=｛x l x∈（M∪N）且x∉（M∩N）｝.（1）：求集合ABA※B（2）：若A∪C=A求实数m组集合
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（1）A=｛0<img class="word-replace" src="/api/getdecpic?picenc=0ad<img class="word-replace" src="/api/getdecpic?picenc=0ad｝B=｛1,2,3｝A∪B={0,1,2,3},A∩B={1,2}所A※B={0,3}（2）A∪C=A所CA真集或C=AC=A符合题意mx=1x解所CA真集则C=｛0｝或C=｛1｝或C=｛2｝或C=空集C=｛0｝应m解C=｛1｝应m值1C=｛2｝应m值1/2C=空集应m值0所实数m组集合{0<img class="word-replace" src="/api/getdecpic?picenc=0ad<img class="word-replace" src="/api/getdecpic?picenc=0ad/2}希望我答所帮助
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其他2条回答
(1)A={0,1,2}；B={1,2,3}；显A∪B={0,1,2,3}A∩B={1,2}故由集合M与N新运算定义A※B={0,3}；(2)若A∪C=A则集合C单元素集{1}或{2}、空集相应m=1或m=1/2或m=0故实数m组集合{0,1/2,1}
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Court did not in court for sentencing.
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SB二代，不过脑子没进水了，因为脑子进水的前提是要有脑子
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出门在外也不愁(x+4)(x-1)-(x+2)(x-3)其中x=负3分之1=（x平方+4x-x-4）-（x平方-3x+2x-6）是用什么方法算的详解 谢谢_百度知道
(x+4)(x-1)-(x+2)(x-3)其中x=负3分之1=（x平方+4x-x-4）-（x平方-3x+2x-6）是用什么方法算的详解 谢谢
项式乘则(a+b)(c+d)=ac+ad+bc+bd
=ac+bc+ad+bd(x+4)(x-1)-(x+2)(x-3)=x&#178;+4x-x-4-(x&#178;-3x+2x-6)=4x+2=4×(-1&#47;3)+2=2&#47;3
其他&2&条热心网友回答
(x+4)(x-1)-(x+2)(x-3)=（x平方+4x-x-4）-（x平方-3x+2x-6）是用是多项式乘法方法算的=（x平方+3x-4)-(x平方-x-6) =4x+2=4×（-1/3）+2=2/3
(x+4)(x-1)-(x+2)(x-3)=（x平方+4x-x-4）-（x平方-3x+2x-6）是用是多项式乘法方法算的=（x平方+3x-4)-(x平方-x-6)=4x+2=4×（-1/3）+2=2/3其中：(x+4)(x-1)=x2+4x-x-4=x2+3x-4，(x+2)(x-3)=x2+2x-3x-6=x2-x-6，最后合并同类项，带入x的值就行本文欢迎转载，转载请注明：转载自中国学网： []
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