High School Mathematics. If the dots p,q , they appear evenly distributed in LPL 3 and LQL 3

Simple axes can be drawn. Draw the desirable area.

It can be simply enumerated, and the eligible ones are (1,1)(1,2)(1,3)(2,1)(2,2)(2,3)(3,1)(3,2)(3,3).

x,y) There are 36 cases, and 9 are eligible.

There is no need to complicate simple questions, if you have to calculate according to the probability formula, it will be a waste of time, In the syllabus of the college entrance examination, the enumeration is also correct.

2p) 2-4(1-q 2)>=0 i.e. p 2+q 2>=1, i.e. find the probability of falling on or outside the circle.

In fact. All fall outside the circle. Because p q is determined by the dice, and |p|<=3,|q|<=3, it is impossible to get 0Then all the cases that can be taken can make the original formula have 2 different real roots.

1. Let the coordinates of p point (x, y), then the vector ap=(x,y-1), the vector pb=(-x,-1-y), the vector pc=(1-x,-y), and according to the conditions, -x 2-y 2+1=kx 2-2kx+k+ky 2. (k+1)x 2+(k+1)y 2-2kx+k-1=0.

x 2 0, so -1 x Shishan 1, and x+k 0, so k -x. Hence k 1.

x 2+2kx+k 2=1-x 2 δ=4k 2+8 0, and the root number of Chaju -2 k annihilated the root number 2.

To sum up, 1 k root number 2.

a<0, the specific steps are as follows: add to the value range of x to take any similar x1, x2, dust and x1>x2, because f(x) is a good code for the subtraction function, so f(socks x1)-f(x2) <0

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If it's an arbitrary triangle, how to determine AC I don't know I'm sorry You can find the length of ab here.

There is by the title.

a+c=2b

bc=a^2

a+b+c=6

The formula 1 and 3 are used to obtain b=2

Then 2c=a2 i.e. 2(4-a)=a2

a^2+2a-8=0

a-2)(a+4)=0

a = 2 or a = -4

a=2 then c=2

a=-4, then c=8

Solution: a,b,c into a series of equal differences, 2b=a+c --1)

Again, b, a, c are proportional sequences, a 2=bc (2) is set by the title, a+b+c=6 (3)

From (3), obtain: a+c=6-b (3').

Replace (3'Generation(1), gets: 2b=6-b.

3b=6, b=2.

a+c=4.

c=4-aa^2=bc

2c.2*(4-a).

a^2+2a-8=0

a+4)(a-2)=0

a+4=0, a1=-4;

a-2=0, a2=2.

2c=a^2=(-4)^2.

2c=16, c1=8, or, 2c=a 2=2 2=4

c2=2.or -- Obviously, this group does not meet the requirements of the question and is discarded.

These three numbers are: a=-4, b=2, c=8

From the series of equal differences, we can know that a+b+c=6

b=2,a+c=4

From b, a, c as the proportional series, we know a*a=b*c

a*a=2c

c=a*a/2

a+c=4, a+a*a2=4, when a=2 or -4a=2, a=b=c=2

A=-4, B=2, C=8

(1) 2 (one in the surface ADC and one in the surface BCD).

2)7 2 You may wish to make the rectangular ABCD a square and make its side length 1, take the AB side as the X axis, take the AD side as the Y axis, and take A as the coordinate origin, and establish a plane Cartesian coordinate system, then P(X, Y), can be directly deduced from the graph When Z 2x +3Y takes the maximum value, P must be on Df, and the linear line equation of Df is X+2Y-2=0, therefore, Z=2X+3-3 2 X=X 2+3, therefore, when X is max 1, Z has a maximum value of 7 2 [Because the time relationship number may be incorrect, but the method is absolutely correct].

3) 4 x 2 + y 2 = 1-z 2 > = 2xy So, s < = 1 (1-z 2)z 2 So, when z 2 = 1-z 2 = 1 2, s takes the maximum value of 4