P 12, I, 1 ~ 5A BCDB II, 6.2 times root number 5/5, 7.60 degrees, 8.2, (root number 2,2 2) III, 9. DF= = root number (MF +DM) = 10 times root number 298, DE.
P2 1，I， 1 ~ 5aCABD II，7。 N/2 (n+2)，8.2n，9。 Let the first three numbers be 16-D, 16, 16+D to form arithmetic progression, and Y =. 25 into a geometric series, (16+D) squared = 16× 25 to get D = 4, so these four numbers are 12, 16, 20, 25.
P30 I， 1 ~ 5c BDD II，6. 13，8. 1/3，9. 15/2
10. Question 1: Let the first term be a 1, the tolerance be d, and A2 2+A3 2 = A4 2+A5 2.
(a5+a3) * (a5-a3)+(a4+a2) * (a4-a2) = 0 because a5-a3=a4-a2=2d=0, a2+a3+a4+a5=0.
Then {4a 1+ 10d=0.
{7a 1+2 1d=7, so a 1=-5d=2.
an=2n-7
sn=n(a 1+an)/2=n(n-6)
Question 2
am * a(m+ 1)/a(m+2)=(2m-7)(2m-5)/(2m-3)= 2m-9+8/2m-3
So 2m-3 can be divisible by 8. 2m-3=-+ 1 m= 1 or 2.
When m =1* a2/a3 =15 = 2 *1-7 = a11.
When m = 2a2 * a3/a4 = 3 = 2 * 5-7 = a5
s(n+ 1)=2(n+ 1)^2+2(n+ 1)
So a (n+1) = s (n+1)-sn = 4n+4.
That is an = 4n(n & gt；; =2)
T(n+ 1)=2-b(n+ 1)
So b (n+1) = bn-b (n+1) = b (n+1) =1/2bn.
And b1= 2-b1= b1=1.
bn=( 1/2)^(n- 1)
So cn = n 2 * (1/2) (n-5)
k=c(n+ 1)/cn=(n+ 1)^2/2n^2
(1- root number 2) < = n
So when n=3, it is the largest, so when n=3, CN+3.
P39 I. 1 ~ 5 BCBDA 6.6，7. 1，2，
9.( 1) When a=0, the original inequality is -x+ 1 < 0, so x >；; 1
(2) when a is not equal to 0, the equation ax 2-(a+1) x+1= 0, so x1=1x 2 =1/a.
If a<0, then x1>; X2 So the solution set of inequality is X.
If a> 1, then x1>; X2, so the solution set is1/a.
10, original formula ((a-2) x-(a-4))/x-2 > 0 when x = 2, x=( 1-2)/(a-2).
When 2 >; (1-2)/(a-2), the solution is x >；; 2 or x
When 2
When a=0, 0 >; 2 empty set, so 1
p47
F (n) = (10+n) (100-80)/radical sign (n+ 1)
The original formula = 1000-80 (root number (n+ 1)+9/ root number (n+ 1)) < =520 and root number (n+ 1)=9/ root number (n+/).
p50
9. If s(x) is a true proposition, then M 2-4.
Any x that belongs to R r(x) is false.
Then sinx+cosx >；; m
= radical 2sin(x+ pie/4) > m
= 2 < = m For all X belongs to the proposition R.r(x) is false, so the root number 2 < = m.
10 、| x+ 1 |+| x-3 | >； eight
x^2-2x- 15=(x-5)(x+3)>； 0
so-3 & gt； X or x> five
x^2+(a-8)x-8a<； = 0a<x< eight
So-3 < a
p53
10.( 1)p equation x 2+x+m = 0 has no real roots. Then 1-4m < 0 is m> 1/4.
(2) the derivative of q function f(x)=xlnx at x=m f' (m) >; 1
If p or q is a false proposition and (1)(2) is false, then m
p56
Question 9 X 2/9-Y 2/ 16c = number of radicals 16+9 = 5 The focus is (+-5,0).
Because |pf 1|*|pf2|=32.
Therefore, cosf1pf2 = (| pf1| 2+| pf2 | 2-| f1F2 | 2)/2 | pf1|||| pf2 |
So | pf1| 2+| pf2 | 2 | pf1| 2 * | pf2 | 2 = 36.
And because | pf1|-pf2 | = 6.
Therefore, (| pf 1 |-| pf2 |) 2 = 36.
|pf 1|^2+|pf2|^2= 100
So cosf 1pf2=0.
So vertical
p59
8. Let the elliptic equation be x 2/a 2+y 2/b 2 =1f (c, 0).
Compress and transform ellipses
x'=x/a，y'=y/b
The ellipse becomes the unit circle x' 2+y' 2 = 1.
F & gtF'(c/a, 0) extends the intersection o to n, the slope of the straight line A 1B2 is known as 1, TM=MO=ON= 1 A 1B2= radical number 2, let T(x.y) be TB2=.
TB2*TA 1=TM*TN
Root number 2 (root number 2x+
Root number 2)= 1*3
X= (radical number 7- 1)/2 (minus, minus)
Y= (radical number 7+ 1)/2
It is known that B 1(0,-1)
(y+ 1)/ (radical number 7+ 1)/2+ 1=(x-0)/ (radical number 7- 1)/2
Let y=0
X=2 (radical number 7)-5
E=c/a=2 Radicals 7-5
The focus is on the x axis,
According to the meaning of the question, 2c=7- 1=6 c=6/2=3.
Long semi-axis a=c+ 1=4
Half axis b= root number (A 2-C 2) = root number 7 The equation is x 2/16+y 2/7 =1.
|op|/|om|=e
Let m(x.y), p(x.k)
|op|= root sign (x 2+k 2)
|om|= radical sign (x 2+y 2)
P | op | = x 2/16+k 2/7 =1on the elliptic circle.
K= root number (1 12-7x 2)/4
x^2+( 1 12-7x^2)/4=e^2(x^2+y^2)
x^2/7( 16e^2-9)/ 16x^2+y^2/7= 1
Ellipse at 3/4 when e>e
p95
8. If K= empty, then x= 1.
If the slope exists, let the straight line be y=k(x- 1)+2, and use the distance from point (0,0) to the straight line to get k=3/4.
So 3x-4y+5=0.
9. (1) Let the major axis of the ellipse be a and the minor axis be a/2. Because the focus is on the axis, the equation can be expressed as x 2/a 2+y 2/(a/2) 2.
If (2, 1) is brought into equation 4/A 2+ 1/A 2/4 = 1, A 2 = 8 a 2/4 = 2, then the equation is X 2/8+Y 2/2 = 1.
(2) the equation of om is y=x/2, so l is parallel to om, so k= 1/2, l=y=x/2+m because x = 0y = m.
Substitute l into the elliptic equation x 2/8+(x/2+m) 2/2 = 1.
2x^2-4mx+m^2-8=0
Because there are two intersections, (-4m) 2-4 * 2 * (m2-8) = 64-8m 2 > 0so-2 radical 2 < m.