Ask for help with a math problem, thank you very much 20

Solution: Set|pn|=x，|pm|=y, angle pan=a, (0so:v=(1 3) *s(triangle mpn) *h(h) (1 3) *x * y * sin(135 degrees)]*2( 2 6) *x * y

2 6) *2sina * 2sin (45 degrees-a degrees) with the formula of product sum difference.

2 3) *cos(2a-45 degrees)-cos(45 degrees)] When 2a-45 degrees = 0 degrees, that is: a = degrees, v takes the maximum value.

In this case, v=(2 3) *1-cos(45 degrees)]=(2 3) *1-(2 2)]=(1 3)*(2-1).

Set |pn|=x，|pm|=y, angle pan=a-degree, (0( 2 6) *2sina * 2sin(45 degrees-a-degree) with the formula of product sum difference.

2 3) *cos(2a-45 degrees)-cos(45 degrees)] When 2a-45 degrees = 0 degrees, that is: a = degrees, v takes the maximum value.

In this case, v=(2 3) *1-cos(45 degrees)]=(2 3) *1-(2 2)]=(1 3)*(2-1).

Use the triangulation induction formula to find the relationship between the tilt angle a+90 and a.

This is very simple, in the circle C, the coordinate of the center c of the circle is (0,0), the radius is R, the point A is (A,B), and the distance to the fixed circle C is equal to the distance to the point A The point coordinate is (X,Y);

then the root number (x 2 + y 2) -r = root number ((x-a) 2 + (y-b) 2)).

Simplification: 4x 2+4y 2=4a 2x 2+4b 2y2+8abxy-4t(ax+by)+t 2 (where t=r 2+a 2+b 2, is a constant).

For the above to be linear, it is necessary to:

4=4a2 and 4=4b2 and 8ab=0

i.e. a=1,-1: b=1,-1 and a=0 (or b=0).

It's impossible.

So 4x 2+4y 2=4a 2x 2+4b 2y2+8abxy-4t(ax+by)+t 2 cannot be a straight line.

The center of the fixed circle C is used as the origin to establish the plane Cartesian coordinate system.

Set the circle c to x +y =r —

The fixed point a is (a, b).

The point at which the distance to both is equal is (w,u).

w=（a+x）/2 ——

u=（b+y）/2 ——

Substitute the formula @&

Get. 2w-a）²+2u-b) ²r ²

So the trajectory is not a straight line.

Further. If a=b=0, i.e., the fixed point a coincides with the center of the fixed circle c then the trajectory is a circle with the origin as the center of the circle and the radius r 2.

If a is not equal to 0 and b is not equal to 0, then the trajectory is an ellipse.

f'(x)=cosx+sinx in (0,3 4) and (7 4,2) derivatives greater than 0 and increasing, in (3 4, 7 4) derivatives less than 0, decreasing. It can be seen that x=3 4 max x=7 4 minimum, with the root number of the original function 2+3 4+1, and the root number 2+7 4+1, depressed, can only lose 100 words.

f`(x)=cosx+sinx+1

Let f(x)=0, i.e. cosx+sinx+1=0

Because (cosx+sinx) 2=1+2sinxcosx, sinxcosx=0 therefore sinx=0, or cosx=0

Because the extremum of the 0 function: f( 2) = 2 + 2 f( ) = 2 + f(3 2) = 3 2

Monotonic interval of the function: (0, 2],[ 2, ]3 2],[3 2,2 )

where in (0, 2],[3 2] increases, and the other two intervals decrease.

The derivative is f(x)'=cosx+sinx+1

2sin (x+45°) +1 under the root number

Let f(x)' 0

Calculated sin(x+45°) root number 2 2

0 x 2 again, then 45° x+45° 2 +45° according to the function graph of sin (due to limited conditions) the maximum value of the function when x=45°, f(x)=root number 2+1, when x=(5 4), the minimum value f(x)=0 of the function is monotonically increasing interval is -(4) x or (3 2) x 2 monotonically decreasing interval is x (3 2) complete.

Is that the title? ：

a=b=when a b=empty set, find the range of values of a.

a=x^2 - 6x + 8 = ( x - 2 )(x - 4 ) 0

Get 2 < x < 4

b=(1) If a=0 then b=empty set, which is in line with the title(2) If a<0 then 3a < x < a < 0 , a b=empty set, which is in line with the title.

3) If a>0 then 0=4, 0=4

In summary, a<=2 3 or a>=4

x 2-6x+8<0 decomposes the factor (x-2)(x-4) <0, and when the solution is 20, a<3a, so the set b=, b is an empty set, and a b is an empty set, which is in line with the title.

3) When a<0, a>3a, so when the set b={xl 3a=4 or a<=0, a b is an empty set.

That is, when a<=2 3 or a>=4, a b is an empty set.

I'm dizzy, that's a question. You're not kidding me, are you? anb=？Is there a solution to this?

It's anb=empty set. There is a solution.

Solution: a={xi20, b={xia=4, so 0=4;

A<0, b={xi3a4, so a<0.

In summary, a<=2 3 or a>=4

a=1.When a=0, b=, ab=, satisfies 2When a≠0, b=, because a b= , then a4 or 3a 2, i.e. a 2 3, so a 2 3 or a 4, and a ≠0

In summary, a (-0) (0,2 3] [4,+

1.Not parallel.

The vectors are parallel, and the horizontal and vertical coordinates correspond to proportional (because vector a is not 0, it must be parallel when it is 0).

2.Where did f(x) come from?

Is it a multiplication of vectors a and b?

Let t=a 1 a, then this inequality is to prove (t 2) 2 t 2, i.e., prove 2 2 t (t 2).

1. If t 0, then the inequality is constant;

2. If t>0, then consider that a 1 a 2 a 2 has t 2. To rationalize the molecule of the inequality to be proved, i.e., to prove: 2 [2 2] 2 [t (t 2), is to prove:

2＋√2≤t＋√(t²－2)。Let f(t)=t (t 2), then the function f(t) is increasing over the interval [2, and since t 2, then the minimum value of f(t) is f(2)=2 (2 2)=2 2, i.e., there is f(t) 2 2.

Thus there are: (a 1 a ) 2 a 1 a 2.

Shirt front He x1=3 x2=1 f(x1 x2)=f(x1)-f(x2),f(3 1)=f(3)-f(1) or send to repent f(1)=0

Let x1=9 x2=3

f(9 3) = f(9)-f(3) So f(9)=-2 and when x>1, f(x) <0, the function is decreasing, f(x)<-2f(x)9

f(1)=f(1)-f(1)=0

f(1/3)=f(1)-f(3)=1

2=-1-1=f(3)-f(1 3)=f(9) Since the cobend is f(x) is the vertical and suffocation function, the solution of f(x)<-2 is Huihe x>9

f(3) =1

f(3)=f(9/3) =f(9) -f(3) =f(9)+1 = 1

So f(9) = 2

Since the function is a positive number of reverberation and repentance, when x>9, f(x)<-2 is solved in Zhaozhou x>9

Solution: Because f(x1 x2) = f(x1)-f(x2), the foci.

f(3/1)=f(3)-f(1)

then, f(1)=0

f(9 3)=f(9)-f(3)=f(3), so the code scrambles f(9)=-2

The function is a decreasing function.

So x>9

I didn't learn. I won't read it, aren't you satisfied with the answers of the people above?

f(3 1)=f(3)=f(3)-f(1), so f(1)=0 and know that the function is a decreasing function, you might as well prepare for f(x)=loga(x) (with a as the base x, you can't fight) then you miss the noise.

f(3)=-1,a=1/3

f(x)<-2, we get x>9

I don't know if it's right or not.

a=2 (x-1) +2x 2>0 (you should be able to see this) and use the ** method.

If y1=2 (x-1) is drawn, then the curve is drawn through (0,1 2) and (1,1) is drawn as y2=-2x 2, then the curve is about the y-axis, and the opening is downward to know that there is no intersection point at this time; Move the parabola upwards to the vertex that coincides with (0,1 2), and there is an intersection point; Moving further up, there are obviously 2 intersections, so a (1 2,+

ps: This is not a quadratic equation and cannot be used with b 2-4ac

2^x-1 +2x^2-a=0

Let's multiply by 2 first.

2^x+4x^2-2a

Two roots. b^2-4ac＞0

That is, 0-4*4*(2 x-2a) 0

i.e. -2 4+2 5a 0

a＞2^(x-1)

Only this is the solution.,I haven't found any other solution for the time being.,If there's any, it's lonely! ~~

a greater than or equal to one-half.

Algorithm: Move A to the right first.

Get 2x squared + 2 x-1 power = a

Extract 2 out.

The square of x + x-2 to the power of x is greater than or equal to 0, so the smallest time is when x=0, and a is equal to 2 times a quarter is equal to one-half. Since x is all numbers, a is greater than or equal to one-half.

Solution: Let x-1=t, then x=t+1

Then 2 t+2(t+1) 2-a=0,2 t=-2(t+1) 2+a=-2t 2-4t-2+a

Let f(x)=2 t, g(x)=-2t 2-4t-2+a (the opening is downward, there is a maximum value).

There are two points of intersection between the images of f(x) and g(x).

and [g(x)] max=g(-1)=a, f(-1)=1 2 To satisfy the topic, there is a>1 (here you can make a simple diagram to understand).