High school math lovers, come, high school math, ask for a great god

This seems to be in the book of the dragon gate.,It's also talked about in class.,ps:"Graceful",Not "full wonderful"。。

Let oa (vector) 1 2, ob (vector) 3 c (x 3+2x+3) (2x 3+x+1), according to the question 1 2< (x 3+2x+3) (2x 3+x+1)<3

Hence it can be said that c is on the vector ab.

That is, c is divided into ab and 0, so the inequality is solved by substituting it according to the formula of the fixed score point, and finally we get x (-5 3) (0, ).

Let the coordinates of the three points m p n on the number axis be x1=1 2,x2=(x 3+2x+3) (2x 3+x+1), x3=3, then.

vector) mn= (vector) mp( >1).

That is, 5 2= [(x 3+2x+3) (2x 3+x+1)-1 2] = [5(2x 3+x+1)] (3x+5)>1 to get the answer x<-5 3 or x>0

The original inequality can be reduced to 0<(3x+5) (2x 3+x+1)<5 (i)If 2x 3+x+1>0, 0<3x+5<10x 3+5x+5 -5<3x<10x 3+5x => x>0 (ii) if 2x 3+x+1<0,0>3x+5>10x 3+5x+5 -5>3x>10x 3+5x => x<-5 3 So x (-5 3) (0, ).

Isn't it also easy to solve the problem by using the number line root method?

Because 1 y 2, the nearest maximum value of y is: 2; The nearest minimum is: 1.

Because 1 x 3, the maximum near x is: 3;The nearest minimum is: 1.

When y is close to the maximum value of 2, z=-x+y has a near maximum: 2-3 and 2-1. The maximum value is 2-1=1, that is, the maximum value of z (the largest value is taken when y is large).

When y is close to the minimum value of 1, z=-x+y has near the minima: 1-3 and 1-1. The minimum value is 1-3, which is the minimum value of z (the smaller value is taken when y is small).

Therefore, the value range of z=-x+y is (1- 3,1).

Solution: The four vertices of the region (x,y) are (1,1)(1,2)( 3,1)( 3,2).

When (1,1), z=-1+1=0;

After (1,2), z=-1+2=1;

After (3,1), z=- 3+1

After (3,2), z=- 3+2

√3+1

The value range of -x is -root 3 to -1 (the inequality sign changes, and the direction of the inequality sign also changes.) So -x y is the direct addition of the same direction inequality, the inequality sign is unchanged, and the numbers are added directly. So the value of z ranges from 1 to 3 to 1.

（1-√3，1）

Equivalent to linear programming.

x (1, 3) y (1,2) draws a region y=x+z known slope to draw a straight line.

z represents the absolute value of upward or downward translation on the image, z units will translate the straight line in the area, and it can be seen that the straight line passes (1,2) and the smallest (3,1) when z is maximum

Substituting z x y

First, find the value range of -x - root number 3 to -1, as long as the inequality sign changes, the direction of the inequality sign changes, so -x y is the direct addition of the same direction inequality, and the number is directly added. The value of z can range from 1-root number 3 to -1. It is important to note that inequalities, change signs, and same-direction inequalities cannot be added and subtracted, and disdirections cannot be operated in this way.

1 x 3, then - 3 -x -1

x min- 3 y min: 1 so z min. = 1- 3 - x max - 1 max 2 so z max = 2 - 1 = 1 so 1 - 3 z 1

<>c will give you a detailed explanation tomorrow The network will be disconnected immediately.

I'll just talk about the method, you do it yourself:

Draw an approximate image of f(x), f(x+a), and f2(x).

The image of f(x) resembles a parabola with an opening upward, f(x+a) is obtained by translating -a from f(x) (the positive and negative properties of a are discussed later), and f2(x) is a similar parabola with a smaller opening than f(x).

After drawing the three function images, it can be seen that f(x+a) and f 2(x) have two intersection points, and in the interval of the two intersection points, f(x+a)>=f 2(x), so as long as [a,a+1] falls in the interval of these two intersection points, f(x+a)>=f 2(x) is always true.

When a>0, find the abscissa of the intersection of f(x+a) and f2(x), so that he satisfies [a,a+1] falls within the interval of these two intersections, and you can see that when a>0, it is not true.

When a<0, find the abscissa of the two intersection points of f(x+a) and f2(x), so that he satisfies [a,a+1] to fall within the interval of these two intersections, and then we can find: a<=-3 4

Draw a formula.,Just pass it **.,You can't enter it.。

Quadratic Functions and Linear Programming Synthesis.

The result of this question is b

(11) Because the alignment is y=-a 4, and the distance from m(3,2) to the alignment is 2+a 4=4, a=8So the intersection coordinates f(0,2), the point p(x,y) is on the line x-y=2, so p(x,x-2), so |pn|=√((x-1)^2+(x-3)^2)=√(2x^2-8x+10), pf|=√(x^2+(x-4)^2)=√(2x^2-8x+16).Therefore, f(x)=(|.)pn|-1)/|pf|=( (2x 2-8x+10)-1) (2x 2-8x+16), commutation, let t=2x 2-8x+8=2(x-2) 2, so t>=0

So the f(t)=( (t+2)-1) (t+8)= ((t+2) (t+8))-1 (t+8)= (1-6 (t+8))-1 (t+8), This formula is preceded by the increasing function of t, and after the negative sign is the decreasing function of t, so f(t) is the increasing function of t, so the minimum value is taken at t=0, there is f(0)= (1 4)-1 8=1 2-1 (2 2)=(2- 2) 4, choose (b) (12) let t=f(x)-lnx, Then there is f(t)=e+1, but f(x)=lnx+t, so f(t)=lnt+t=e+1, because lnt+t is a monotonic increasing function, so t=e, so f(x)=lnx+eSo f'(x)=1/x, g(x)=f(x)-f'(x)=lnx-1 x+e, because g'(x)=1 x+1 x 2>0(x>0), so g(x) increases monotonically, and g(1 e)=-1-e+e=-10, so g(x) has only one zero point in the defined domain.

Because the PD is perpendicular to ABCD, the PD is perpendicular to AD;

Because ABCD is rectangular, AD is perpendicular to DC;

Therefore AD perpendicular plane PDC, so AD perpendicular PF.

Connect to AC and hand over to O. The triangle AOD is similar to COE, so OC OA=CE AD=1 2; and fc fp=1 2, so oc oa=fc fp; The triangle COF is similar to the cap, ap of, so the ap plane def

PF1 = PF2 + 2A = 9 2 + 2A because PF2 F1F2 so PF2 +F1F2 = PF1 and F1F2 = 2C = 2A multiplied by the eccentricity. substitution yields a=6, so c=3 7; b=3√3;Dashed axis = 2b = 6 3

2⑤(x-1)²+y-1)²=1

(1) Find the red ball first, and leave 2 red and 2 black, with a probability of 2 4 = 1 2 =

2) Put it back, there are 3 red and 2 black, and the probability is 3 5=

The first is multiplication, and the second is multiplication.

1/5 1/4 = 1/20, 2nd 25/25.