Two. 1.- 2 2.3.4. 1

Answer: 1. 1. According to the conditions, a2-3a+ 1=0, b2-3b+ 1=0, a≠b,

So A and B are the two roots of the unary quadratic equation x2-3x+ 1=0, so a+b=3 and ab= 1.

Therefore += = = 7.

2. Because AD, BE and CF are the three heights of △ABC, it is easy to know the four * * * cycles of B, C, E and F,

So △AEF∽△ABC, so = =, that is, cos∠BAC=, so sin∠BAC=.

In Rt△ABE, BE=ABsin∠BAC=6×=.

3. Two digits can be 12, 13, 14, 15, 2 1, 23, 24, 25, 3 1, 32, 34, 35. 15,21,24,42,45,51,54, * * 8, so the probability that this number is a multiple of 3 is =.

4. ∠ABC =12, BM is the bisector of ∠ ABC, ∠ MBC = (180-12) = 84.

∠BCM = 180-∠ACB = 180- 132 = 48，∴ BCM = 180-84-48。

∴BM=BC and ∠ ACN = (180-∠ ACB) = (180-132) = 24,

∴∠bnc= 180-∠ABC-∠BCN = 180- 12-(∠AC b+∠can)= 12 =∠ABC。

∴CN=CB. Therefore, BM=BC=CN.

It is easy to know that the prices of five kinds of goods can be different after 4 days.

Let the prices of five commodities before price reduction be a and n days later, the price of each commodity after n days can be expressed as a (1-10%) k (1-20%) n-k = a () k () n-k, where k is a natural number. a ()i+ 1 ()n-i- 1，a ()i+2 ()n-i-2，a ()i+3 ()n-i-3，a ()i+4 ()n-i-4。

Where I is a natural number not exceeding n, so the minimum value of R is =()4.

6.∫(x-)(y-)= 2008，

∴x-==y+,y-==x+.

X=y can be obtained from the above two formulas, so (x-)2=2008. The solution is x2=2008.

So 3x2-2Y2+3x-3y-2007 = 3x2-2x2+3x-3x-2007 = x2-2007 =1.

2. 1.∫a2 =()2 = = 1-a，∴a2+a= 1.

∴ Original formula =

= = =-=--( 1+a+a2)=-( 1+ 1)=-2。

2. Let the midpoint of BD be O and connected with AO, then AO⊥BD, AO=OB=, MO==,

∴MB=MO-OB=. and ∠ ABM = ∠ NDA = 135,

∠NAD =∠MAN-∠DAB-∠MAB = 135-90-∠MAB = 45-∠MAB =∠AMB，

So △ADN∽△MBA, so =, so DN = Ba =× 1 =. According to symmetry,

The area of the quadrangle AMCN is s = 2s△ man = 2××× Mn× ao = 2××× (++)× = )× =.

3. According to the meaning of the question, M and N are two roots of the unary quadratic equation x2+ax+b=0, so m+n=-a and Mn = B. 。

∵|m|+|n|≤ 1,∴|m+n|≤|m|+|n|≤ 1,|m-n|≤|m|+|n|≤ 1.

Equation x2+ax+b=0 △=a2-4b≥0 discriminant, ∴b≤=≤.

4b = 4mn = (m+n) 2-(m-n) 2 ≥1-(m-n) 2 ≥-1,so b≥- and the equal sign are obtained when m=-n=;

4b = 4mn = (m+n) 2-(m-n) 2 ≤1-(m-n) 2 ≤1,so b≤, when m=n=, an equal sign is obtained. So p=, q=-, and then | p || q.

4. 12 to 32, the result only accounts for 1 digit, and * * * accounts for 1×3=3 digits; 42 to 92, the results only account for 2 digits, and * * * accounts for 2×6= 12 digits; From 102 to 3 12, the result only takes 3 digits, and * * * takes 3×22=66 digits; From 322 to 992, the results only occupy 4 digits each, and * * * occupies 4×68=272 digits; From 1002 to 3 162, the result only takes 5 digits, and * * takes 5×2 17= 1085 digits; There is a gap of 2008-(3+12+66+272+1085) = 570 digits. 3 172 to 4 1 12, and the results only account for 6 digits, and * * accounts for 6× 95 =