For the function f (x), if there exists x0 ∈ r such that f (x0) = x0 holds, then x0 is called the fixed point of F (x) For the function f (x), if there exists x0 ∈ r such that f (x0) = x0 holds, then x0 is called the fixed point of F (x), if f (x) = x ^ 2 + A / (bx-c) (B, C ∈ n +) There are only two fixed points 0,2, and f (- 2)

If 0 is a fixed point of F (x), we have:
f(0)=a/(-c)=0,a=0 f(x)=x^2
But f (2) = 2 ^ 2 = 42, 2 is not the fixed point of F (x)

For function f (x), if x0 belongs to R and f (x0) = x0, then x0 is called the fixed point of function f (x). We know that f (x) = ax ^ 2 = (B + 1) x + (B-1) (a is not equal to 0)
(1) When a = 1, B = - 2, find the fixed point of F (x)
(2) If the function f (x) has two different fixed points for any real number, the value range of a is obtained
(3) Under the condition of (2), if the abscissa of two points a and B on the image of y = f (x) is the fixed point of F (x), and the two points a and B are symmetric with respect to the straight line y = KX + 1 / (2a & # 178; + 1), the minimum value of B is obtained

I do it according to f = ax ^ 2 + (B + 1) x + (B-1). (1) omit (2) the function f has two fixed points, that is, f = x has two ax ^ 2 + (B + 1) x + (B-1) = x △ x ＞ 0 constant, B ^ 2-4ab + 4a) > 0 constant △ x ＜ 0 constant, the solution is (0,1) (3) first k = - 1, let a (x1, x1), B (X2, x2), then X1 + x2 =

It is known that the set a = {x | 1 ≤ x ≤ 4}, f (x) = x ^ 2 + PX + Q and G (x) = x + 4 / X are functions defined on a, and obtain the minimum value at x0 and satisfy f (x0) = g (x0), so as to find the maximum value of F (x) on a

g'(x)=1-4/(x^2)
g(1)=5
g(4)=5
When G '(x) = 0, x = 2
G (2) = 4 is the minimum
So x0 = 2
So f (x) is the minimum when x = 2
So f '(2) = 0
f'(2)=2*2+p=0
p=-4
Because f (x0) = g (x0)
So f (x0) = 4
2*2-2*4+q=4
q=8
f(x)=x^2-4*x+8
f(1)=5
f(4)=8
The function is continuous and f '(2) = 0
So the maximum value is f (4) = 8

As shown in the figure, the bottom radius of a cone is 2, the length of bus Pb is 6, and D is the midpoint of Pb. An ant starts from point a and crawls along the side of the cone to point D, then the shortest distance for the ant to crawl is ()
A. 3B. 23C. 33D. 3

Let the center angle of the fan-shaped circle after the expansion of the side face of the cone be n ° and 4 π = n π × 6180 according to the fact that the circumference of the bottom face is equal to the arc length of the fan-shaped circle after the expansion. The solution is n = 120 ° so ∠ APD = 120 °△ 2 = 60 ° in the expansion diagram. Because the radius PA = Pb, ∠ APB = 60 °, the triangle PAB is an equilateral triangle, and ∵ D is the midpoint of Pb, so a is an equilateral triangle D ⊥ Pb, in the right triangle pad, PA = 6, PD = 3, according to the Pythagorean theorem, we get ad = 33, so the shortest distance for ants to crawl is 33

Simplification of mathematical absolute value
X-|X+2/1|-2/1

x-x-2-2=-4

The diameter of circle a is 8 cm, 1 / 7 more than the radius of the hospital. The area of circle B is () square cm, and the perimeter of circle B is () cm

The diameter of circle a is 8 cm, 1 / 7 more than the radius of circle B,
Let the radius of circle B be X
X (1 + 1 / 7) = 8, the value after the radius of B circle is increased by 1 / 7 is equal to 8
So the area is 7 ^ 2 * 3.14 = 49 * 3.14, and this 7 ^ 2 is the square of 7
Perimeter 2 * 7 * 3.14 = 14 * 3.14
It's 1 / 7 more than the radius of circle B. I can't tell what it means. What I understand is the one with the best number
This 1 / 7 is not marked cm, it should be a matter of proportion

The shape of mpnq is judged and proved by taking AB and AC as triangle ABC, making square ABDE and ACGF, and m, N, P and Q as the midpoint of EF, BC, EB and PC

Diamond
[brief analysis]
Connecting BF, CE
Easy syndrome MP / / BF, NQ / / BF
So MP / / NQ
Similarly, MQ / / CE / / PN
Therefore, mpnq is a parallelogram
△CAE≌△FAB
∴CE=FB
∵MP=1/2·BF
MQ=1/2·CE
∴ MP=MQ
The mpnq is a diamond

The chord length is 11.7m, and the distance from chord to arc is 3M. What is the area of this arc?

R=H/2+L^2/(8*H)
=3/2+11.7^2/(8*3)
=7.204m
A=2*ARC SIN((L/2)/R)
=2*ARC SIN((11.7/2)/7.204)
=108.6 degrees
S=PI*R^2*A/360-L*R*COS(A/2)/2
=PI*7.204^2*108.6/360-11.7*7.204*COS(108.6/2)/2
=588 square meters

The summation problem of the sequence of higher one
1. The first term of the equal ratio sequence is a, the common ratio is Q, Sn is the sum of the first n terms, S1 + S2 + +Sn
2. The general term formula of sequence {an} is an = 1 / [√ n + √ (n + 1)]. If the sum of the first n terms is 10, then the number of terms n is?

1.
an=aq^(n-1)
Sn=a(q^n-1)/(q-1)
S1+S2+… +Sn
=[a(q^1-1)/(q-1)]+[a(q^2-1)/(q-1)]+[a(q^3-1)/(q-1)]+…… +{a[q^(n-1)-1]/(q-1)}+[a(q^n-1)/(q-1)]
=[a/(q-1)]{(q^1-1)+(q^2-1)+(q^3-1)+…… +[q^(n-1)-1]+(q^n-1)}
=[a/(q-1)][q^1+q^2+q^3+…… +q^(n-1)+q^n-n]
=[a/(q-1)][q(q^n-1)/(q-1)-n]
=aq(q^n-1)/(q-1)^2-na/(q-1)
two
an=1/[√n+√(n+1)]
=[-√n+√(n+1)]/{[√n+√(n+1)][-√n+√(n+1)]}
=-√n+√(n+1)
an=-√n+√(n+1)
a(n-1)=-√(n-1)+√n
a(n-2)=-√(n-2)+√(n-1)
……
a3=-√3+√4
a2=-√2+√3
a1=-√1+√2
Add two
sn=√(n+1)-1
sn=√(n+1)-1=10
√(n+1)=11
n+1=121
n=120

What is the poem about berthing a boat in Guazhou?

There is only a few mountains between the waters of Guazhou in Jingkou. The spring breeze is green on the South Bank of the river. When will the moon shine on me Wang Anshi's