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The problem of mathematics score in the second volume of the seventh grade
1. Xiaoming answered incorrectly. (a+3)? /(a-3)(a+3)=(a+3)/(a-3) This simplified step is wrong, because it can only be eliminated when a+3≠0. When (a+3)*(a-3)=0, the score is meaningless.

Therefore, when a=3 or a=-3, the score is meaningless.

2. To put it simply, since I said that x/y=2, I will assume that x=2, y= 1, and substitute the available answer =3 (this kind of fill-in-the-blank question or multiple-choice question can be calculated like this, which is fast and time-saving).

3.[( 1+4)/(a? -4)]×[(a+2)/a]= [a? /(a? -4)] [ (a+2)/a ] =a? (a+2)/[(a+2)(a-2)a]=a/(a-2), and a≠0, a≠2, a≠-2,

4.(x? -9)/(x? -3x) = (x+3) (x-3)/[x (x-3)] = (x+3)/x =1+3/x, and x≠3, x≠0.