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It is calculated using the formula for the distance between two points.

ac^2=100

ab 2 = 25 (this is very basic, no more talking, mainly below) let the bisector of a cross ab to m

m is on bc, so it is necessary to find the bc equation.

Find bc:y=-2x 11+69 11 (this is also very basic, be careful not to make mistakes).

So m can be set to (m, -2m 11+69 11) according to the angular bisector theorem.

mc/mb=ac/ab

So mc 2 mb 2=ac 2 ab 2=100 25=4 is expressed by the formula for the distance between two points to get the equation mc,mb.

m+4)^2+(-2m/11+69/11-7)^2]/[(m-7)^2+(-2m/11+69/11-5)^2]=4

This equation is very difficult to solve, and it is a lot of calculations, so be prepared and do it head-on (I recommend that you start with a score and get rid of a 121, which is easier to do).

Finally, 3m 2-64m + 180=0 is obtained

The solution yields m = 18 or 10 3

Obviously, m=18 is too far away and off-topic.

So m=10 3

So m(10 3, 17 3).

The equation for the angular bisector am found at two points of simultaneous a,m is .

y=-7x+29

7x+y-29=0

Although I didn't calculate this question myself, I also pressed a lot of calculators... Give us a little bit ...

ab：4x-3y-13=0

ac: 3x+4y-16=0

Let the bisector of angle a be a(x-4)+b(y-1)=0 (pass through point a) and the distance from the point on the line to ab is equal for ac.

i.e. 4x-3y-13=3x+4y-16

The system of equations ax+by-4a-b=0 and x-7y+3=0a 1=b -7=(-4a-b) 3 is obtained

The solution gives b=-7a, so x-7y+3=0 is the equation for the bisector of the angle a.

The linear equation for ab is l1:4x-3y-13=0, the linear equation for ac is l2:3x+4y-16=0, and the linear equation for bc is l3:

2x+11y-69=0 Let the angular bisector of a be crossed by AD at D, D(X, (69-2X) 11), and the distance from D to AB is, D1=|4x+3*(2x-69)/11-13|/√（4*4+3*3）=|4x+3*(2x-69)/11-13|/5

The distance from D to AC is d2=|3x+4*(69-2x)/11-16|/√(3*3+4*4)=|3x+4*(69-2x)/11-16|5,AD is the bisector of angle a, so the distance from d to ab,ac is equal, i.e. d1=d2,|4x+3*(2x-69)/11-13|/5=|3x+4*(69-2x)/11-16|5. Solve the equation to get x = 18 or 10 3, because d is on bc, so -4< x<7, so x=10 3,d(10 3,17 3), so the equation for ad is, 7x+y-29=0, which is the equation for the bisector of the angle a.

Use the distance from the point on the angle bisector to the two sides of the angle to be equal, ab:4x-3y-13=0 ac:3x+4y-16=0 Let the coordinates of any point of the angle bisector be (x,y) Use the distance formula from the point to the line to get 3x+4y-16=4x-3y-13 or 3x+4y-16=-(4x-3y-13) and bring a(4,1) into the test.

The easy way.

1.Let the midpoint of the line segment bc be d, and d(,6) can be known

2.From the coordinates of the two points of AD, the slope of the equation is -2

3.So from the point a and the slope, the equation is 2x+y-9=0

1.Ask for a derivation. f'(x)=1-lnx/x² f'(x) x e at 0 so (0,e) is the increasing interval (e,+ is the decreasing interval.

2.（0，-√3），（0.3) is the focus.

The sum of distances is a fixed value, indicating that it may be an ellipse.

So c=- 3 b=- 2 -3=1 a=2

So the curve is an ellipse x 4 +y 3=1

Probability of being detonated = 1 - Probability of not being able to detonate = 1 - Probability of not hitting all of them - Probability of hitting exactly 1 shot.

The one multiplied by 5 is the number of hits.

2) Number of shots = x (2< = x <=4) means that the x-th shot hits and the first x-1 shot also hits 1 time.

So when x<5 the probability is 2 3 * x-1) *2 3 * 1 3) x-2).

x=5, because the shooting must stop, so whether it hits or not is the same.

The probability is the sum of the probabilities of 1 minus the number of shots = 2,3,4.

（1）f'(x)=1 lnx Because when x=1 lnx=0 and lnx monotonically increases, so when x is greater than 1 it is an increasing function, and because the denominator is not 0, so f(x) monotonically increases in the interval (0, $2), because the distance to p is constant, so c is an ellipse, and because the intersection is on the y-axis c = root number 3, a=2, b 2=a 2-b 2=1, the ellipse c = (y 2) 4 (x 2) = 1

Question 1 is a direct derivation, and Question 2 is an ellipse at a glance.

The low radius of the cylinder is 5 for the busbar and the shortest is half a cylinder (I don't know?). ) is as short as [52+(.]

The straight line between two points is the shortest. Place the sides of the cylinder, getting.

l=√[5²+（5π／2）²]

Put the side, easy to know, for ((approx. equal to.)

Solution 1, log2(a x-1)>1

then a x>3=a [log2(a 3)].

When 01, x>log2(a3).

2. I don't understand the meaning of the question and can't answer.

3、（1）f(2)+f(-2)=0

2) f(x) is an odd function defined on r, then let x<=0 and -x>=0

f(-x)=a^(-x)-1=-f(x)

So, when x<=0, f(x)=1-a(-x), and when x>=0, f(x)=a x-1

3) When x-1<=0, i.e., x<=1, -11, 1-x1-loga(2) can be obtained

4. According to the meaning of the topic, it can be obtained: (4m 2+1 m 2+1) x 2-2x+5>=0

For x belongs to [2 3, positive infinity], the above inequalities hold.

then the axis of symmetry x=1 (4m 2+1 m 2+1)<=2 3, and (4m 2+1 m 2+1)*4 9-2*2 3+5>=0

Joint solution: the range of m is OK.

i.e. (x-a) (x-1) < 0

The zero points are a and 1

If a<1

then A11 is a minimum of 2

So the maximum is x=7

Then 7

x²-（a+1）x+a＜0

x-a)(x-1)<0

If a<1, the solution is: a1, and the solution is: 11

The solution is: 1 The sum of the integer solutions is 27, and the result is 7 respectively

(x-a)(x-1)<0

a>1

Let the largest integer solution be t

Then there is (2+t)(t-1) 2=27 (the formula for summing the difference series) to get t=7

x t7 again

x - (a+1)x+a=(x-a)(x-1)<0, because the sum of all integer solutions is 27, so a>1 and the solution set is 1

This question does not need a master to answer, it is discussed in two categories, and the number and form are combined:

1) When a>1, the function y=|a^x-1|(a 0, and a≠1), from negative infinity to 0, and y from 1 to 0; from 0 to positive infinity, y from 0 to positive infinity;

To have two common points with the line y=2a, it can only be 0 (note that the line y=2a is a line parallel to the x-axis, which is different from y=2x) (2) when 0

a belongs to (0, 1, 2).

The image of the function y=a x-1 is made first. Then add the absolute value, turn the part below the x-axis up, there are two intersection points, at a glance.

Divide a 1 and a = 1 and a 1 respectively as images Pay attention to the absolute value.