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