I'm not sure about the specific test paper. According to the information you provided, I can only find the "Mathematics Model Two Examination Paper for Senior High School Entrance Examination in Baoan District, Shenzhen City, Guangdong Province" on the Internet. Accordingly, I provided answers to fill in the blanks:

13 . x(x-2)/(x-2)(x-2)= x/x-2

14. The fine word selected from package A is 1/2, and the heart word selected from package B is 1/2, so the probability of selection is1/2 *1/2 =1/4.

15. Because y=x, this angle is 45 degrees, that is, the vertical line is equal to the horizontal line (for example, OA1= b1), OA2=B2A2 ... and the number of roots ob1=/kloc-. So OA2=OB 1= root number 1 is also calculated as follows: OA3= root number 4a4 = root number 8a5 = root number 16 ... So the data are listed: the square of OA65438+0 = the square of 65438+0a2 = the square of 2oa3. 6 ... So it is concluded that the square of the n- 1 power of OAn =2 is OAn = the root sign (the n-65438 power of 2).

16. Documents D and C found online should be called. Solution: let BC' and AD intersect at n, EF and AD intersect at m,

According to the nature of folding, we can get: ∠NBD=∠CBD, AM=DM= AD, ∠ FMD = ∠ EMD = 90.

∵ quadrilateral ABCD is a rectangle,

∴AD∥BC,AD=BC=4,∠BAD=90，

∴∠ADB=∠CBD，

∴∠NBD=∠ADB，

∴BN=DN，

Let AN=x, then BN=DN=4-x,

At Rt△ABN, AB2+AN2=BN2,

∴32+x square =(4-x) square,

∴x=7/8，

That is, AN=7/8,

∫C′D = CD = AB = 3，∠BAD =∠C′= 90，∠ANB =∠C′ND，

∴△anb≌△c′nd(aas)，

∴∠FDM=∠ABN，

∴tan∠FDM=tan∠ABN，

∴ AN/AB=MF/MD，

∴ 7/8 /3=MF/2，

∴MF=7/ 12，

According to the nature of the fold, EF⊥AD,

∴EF∥AB，

AM = DM，

∴ME= 1/2AB=3/2

∴ef=me+mf=3/2+7/ 12=25/ 12