A high school math problem! Add 20 points to it! Words count!!

1) Let p=(x1,x1 squared -1).

q = (x2, x2 squared -1).

k1 = 2 * x1 = (x1 squared + 1) (x1-a) i.e. 2 * x1 * (x1-a) = (x1 squared + 1) k2 = 2 * x2 = (x2 squared + 1) (x2-a) i.e. 2 * x2 * (x2-a) = (x2 square + 1), that is, x1 and x2 are equations.

2*x*(x-a) = (square of x + 1), so x1*x2=-1 x1+x2=2a (the arrangement of the equation will be, and then Vida's formula).

k1*k2=4x1*x2=-4

2) From the set P and Q coordinates, the slope of Pq can be obtained as:

x1 squared - x2 squared) (x1-x2)=(x1+x2) Let the equation for pq be: y=(x1+x2)x+b

Bringing in the coordinates of point p gives b=1-x1*x2=2y=2ax+2 through the point (0,2).

apq/|pq|That is, the distance from the point a(a,0) to pq, set to ab, then the slope of ab is (1 2a), and then through the point a, so the function expression is y=(1 2a), and the simultaneous ab, pq function expression, with the coordinates of point b, can be obtained.

ab|= square of {(4a squared + 4a) squared + [1 + (4*a squared + 5a)] under the root number.

It must be the shortest when a=0 ......

In this way, k1 and k2 must be opposite numbers, so k1=2 and then pass (0,2) by pq, you can know ap=aq=root number three, you cross it down......It's annoying to know that you can't directly type characters with mathtype or something......

Question 1 because the length of the line between the 4 points is .

2 then the ball is the outside ball of the ABCD tetrahedron.

And according to the center of the outside ball. Tetrahedron.

a (a is the side length).

The radius of the ball is .

The distance from point O to the surface BCD is D

The perpendicular foot is P because the triangle BCD is a regular triangle, so there is.

pd=pc=pb=2√

3 3 and then according to the Pythagorean theorem there is d=

For the second question, choose B, if the vector BC+vector BA=2 vector BP, according to the parallelogram rule, then P is at the midpoint of BC, then it is easy to prove that option B is true.

The first question is questionable, do these four points form a square on the same plane, or do these four points form a tetrahedron?

(1)a2=5/8,a3=15/32

2) bn = 1+24an an=(bn 2-1) 24, substitute an+1=1 16(1+4an+ 1+24an), and continuously simplify to get 4b(n+1) 2=bn 2+6bn+9=(bn+3) 2, and bn= 1+24an>0

2b(n+1)=bn+3,2[b（n+1）-3]=bn-3

Therefore, [b(n+1)-3] (bn-3)=1 2 and the first term (b1-3)=2 is not 0, so it is a proportional series.

bn=(1/2）^(n-2)+3

3) The general term of an can be deduced from the bn general term of the second question.

Then we get f(n)=(7-bn)*(bn-3) 16=1-(1 2) 2n

So f(1)·f(2)·f(n)=( This form is a bit like a 2nd term, as if some formula is used, (>1 2 is exactly that.

That's all I can do for you.

Alas, mathematics has degraded a lot.

Question 1 The difference between the fast clock and the slow clock per hour is 4 plus 1 minute, which is 5 minutes, and now the difference between the fast clock and the slow clock is 1 hour, that is, 60 minutes divided by 5 equals 12 hours, and the fast clock goes 61 times 12, which is 732 minutes, and now it is 11 o'clock, which means that the standard time to start is 11 o'clock minus 732 minutes, which is 22:48 minutes yesterday, and the current standard time is 22:48 yesterday plus 12 hours, that is, 10:48 today, a bit long, please forgive me I hope you understand.

1. The difference between fast and slow bells is one hour, that is, 60 minutes, the difference between fast and full bells is 5 minutes, 60 5 = 12 minutes, and the standard time is 10:48

2. I didn't understand.

1.Set the standard time to be x minutes, the fast clock display time is x(1+1 60), and the slow clock time is x(1-4 60).

The difference between the display time of the fast clock and the slow clock is 60 minutes, that is, x(1+1 60)-x(1-4 60)=60, and x=720 is obtained, and the display time of the fast clock is 732, which is 12 minutes faster than the standard time, and the standard time is 10:48 minutes.

2.Let the bamboo pole length l, from the question 2x-l=20%l, get l=x, from y:x=2, get l=5 6y or 5 3x

Question 2: Choose A. set, the rod length is l meters. l is less than y, y=2x, so l is less than x. Solved from the problem 2x=(1+20%)l: l=5 3x

Fast clocks travel 5 minutes more per hour than slow clocks, and fast clocks now travel 60 minutes longer than slow clocks, indicating that they have traveled 12 hours from the adjusted standard time (60 5 = 12). Because the clock travels 12 minutes longer than the standard time in 12 hours, the standard time at this time is 11:00-0:

Let the rod length be z, then z=2x, known y:x=2, z y=2x. So.

z=2x-0,.2z ,z=2/

D should be chosen

f=(-6,4)

1) e=[4, positive infinity) (negative infinity, -2].

e cross f=(-6,-2].

2) e=[m+1, positive infinity) (negative infinity, 1-m]e f=r, so 1-m>=-6 and 1+m<=4 get m<=3

Change it up a bit. e=,f=

e=, the range plot is drawn below.

e∩f=｛x∣x<=-2｝

e∪f=re=,f=

m belongs to (-infinity, 3).

Left side of the equation = (lg4 lg3) * (lg8 lg4) * (lgm lg8) = lgm lg3 = log3(m), that is, the logarithm with 3 as the base m.

The logarithms in the above equation are all based on 10 except the last one.

Right side of the equation = so m= 3