SOS math problems are so difficult, SOS high school math difficult .

1 It is observed that in order to use the conclusion, two values must be taken to make f(x) the same, because f(x) is the multiplication of 2 factors, and it is natural to think that any factor is 0, and f(x) is 0, and in connection with the conclusion, we know that x1 takes 2, x2 takes pai (2pai is larger), (2, pai) (1, 4), f(2) = f(pai), so there is c (1, 4), f'(c) 0

2 This kind of problem should construct a functional proof, and equations like this are generally constructed in relation to e x, because of its derivative function, let f(x) = e xf(x), f(a) = f(b) 0, so f(a) = f(b) 0, so there is c (a, b), f'(c) 0, i.e., e c(f(c)+f'(c)) 0 and e c>0, so f(c)+f'(c)=0

In fact, if we take f(x)=e -xf(x), then f(c) f can be proved'(c)=0...

3 The known proposition is called Roll's theorem, which is taught in the unary calculus of the freshman year, and is called the Lagrange median theorem, that is, the closed interval is continuous and the open interval is derivable, then f(b)-f(a)=f'(c) (b-a), c belongs to (a, b), and it is also a constructor certificate, and it is difficult to think of a new function, so if the landlord has not learned, this question can be given up, after all, it is more difficult to guess.

by f(2) = f( ) = 0

f(x) is derivable in (2, ) and [2, ] continuous.

Obtain: the presence of c (2, )1,4) makes f'(c)=0, let g(x)=f(x)e x

So g(a)=g(b)=0

Derivable by g(x) in (a,b), and continuously obtained on [a,b]: there is c

Make g'(c)=f(c)e^c+f'(c)e^c=[f(c)+f'(c)]e^c=0

Obviously e c is not zero, then f(c) + f'(c)=0 Lagrangian theorem: If the function y=f(x) is derivable in (a,b), it is continuous (the image is uninterrupted) on [a,b].Well, there is at least one c (a, b) such that f'(c)=((f(b)-f(a))/(b-a)).

Both propositions are called the differential median theorem, while Roll's theorem is a special case of the Lagrangian theorem.

f(2) = f(π)=0

f(x) is derivable in (2, ) and [2, ] continuous.

There is c(2, )1,4), so that f'(c)=0g(x)=xf(x)

g(a)=g(b)=0

g(x) is derivable within (a,b) and continuous on [a,b].

CG is present'(c)=f(c)+cf'(c)=0

If the function y=f(x) is derivable in (a,b), it is continuous (the image is uninterrupted) on [a,b].Well, there is at least one c (a, b) such that f'(c)=((f(b)-f(a))/(b-a)).

The first question is easy, f(0)=0, f(2)=0, and there is f on (0,2).'(c)=0

1,4) belongs to (0, ).

This type of problem should be solved using substitution, because the function is independent of the letter represented.

Replace x with u, let u=x+1, then x=u-1

It is obtained from the original formula f(x+1)=3x +8x+6.

f（u）=3(u-1)²+8(u-1)+6=3u²-6u+3+8u-8+6

3u²+2u+1

Replace u with x, f(x) = 3x + 2x + 1

In addition, y=x+1 is brought into f(x+1)=3x+8x+6, then x=y-1, that is, f(y)=3(y-1) +8(y-1)+6 is simplified to obtain f(y)=3y+2y+1

Replace y with x, i.e. f(x) = 3x + 2x + 1

Let x+1=t, x=t-1, bring in, f(t)=3(t-1) 2+8(t-1)+6, i.e. f(t)=3t 2+2t+1, so the answer is 3x 2 + 2x + 1

18. In fact, it is to use OC and OB perpendicular, BC and OA to be parallel, and let C(X, Y) to obtain the system of equations.

2y-x=0

x+1/y-2=3/1

Solution. x=14 y=7

So AC = (14-3, 7-1) = (11, 6) and because OD+OA=OC

So od=ac=(11,6).

It's been a long time since I've done a vector problem, and it's not guaranteed.

19、（1）y=p*x-(1/10x²+5x+1000)=(1/b-1/10)x²+(a-5)x-1000

2）y'=2(1 b-1 10)x+a-5 from the question x=150 when y'=0，p=40

300（1/b-1/10）+a-5=0

a+150/b=40

Solve the system of equations.

a=-35b=30/7

Find the two vectors as (a,1) and (4,a) respectively, and then compare the value of a square equal to 4, and solve a is equal to 2 or -2, because the same direction is parallel and parallel, so a = 2

Solution: Make NF BC to F

In the right-angle BEC and right-angle FMN, b= nfm=90°, in RT BEC and RT FMN, CE=Mn BC=FN and BEC FMN

mnf=∠mce=35°

anm=90°-∠mnf=55°

So the answer is: 55°

It was supposed to be just a square. Don't figure it.,,But it came out again ce or something.。。 So it's a must!

We need a picture!!