The function of the first and second degree of the higher one, and the function of the first and sec

The two real roots of x2+2mx-m+12=0 are greater than 2, first of all, to ensure that the tower of Del is 0

4m^2+4m-48》0

m^2+m-12》0

m+4)(m-3)》0

m" 3, or m "-4

The two are x1 and x2

x1+x2=-2m>4 => m<-2

x1x2=-m+12>4 => m<8

Therefore, the value range of m is m《-4

If there is a real root, the discriminant formula is greater than or equal to 0

So 4m 2 + 4m - 12 > = 0

4(m-3)(m+4)>=0

m>=3,m<=-4

x1+x2=-2m,x1x2=-m+12

x1>2,x2>2

x1-2>0,x2-2>0

So (x1-2) + (x2-2) >0

x1-2)(x2-2)>0

x1-2)+(x2-2)>0

x1+x2>4

2m>4

m<-2

x1-2)(x2-2)>0

x1x2-2(x1+x2)+4>0

m+12+4m+4>0

m>-16/3

So-16 3

According to the function image, find :

The group of inequalities is derived from the meaning of the title.

2m) 2-4*1*(-m+12)>=0 has solid roots.

x0=-2m 2=-m>2 The axis of symmetry is greater than 2

f(2)=2 2+4m-m+12>0 The value of the function at x=2 is greater than zero.

Solution. 16/3

The f(x)=x -ax+a 2=(x-a 2) -a 4+a 2 function and the coincidental image are parabolas with the opening pointing upwards and the axis of symmetry x=a2.

When 0g(a) = -a 4 + a 2

When a 2, when x=1, f(x) is the smallest, then the width is only.

g(a)=1-a/2

When 0g(a)=-a 4+a 2=-(1 4)(a -2a+1)+1 4=-(1 4)(a-1) +1 4

At this point, when a=1, g(a) is the largest and the value is 1 4 when a2.

g(a)=1-a/2≤0

At this time, when a=2, g(a) is the largest and the value is 0, and the maximum value of g(a) is 1 4

Solution: (1) is given by the relation f(3+x)=f(3-x)f(x) is a function about x=3 symmetry (you can see it by drawing the graph, combined with the graph to solve the problem), so x=3 is the midpoint of x1, x2, so there are (x1+x2) 2=3, and x1+x2=6; (Graphical incorporation is a great way to solve problems).

x1+x2=6

Solve it first with algebraic methods:

f(3+x)=f(3-x)

f（3+x-3）=f（3-x+3）

f（x）=f（6-x）

f（x1）=f（6-x1）

Since the two real roots satisfy f(x1)=0=f(x2), x2=6-x1

What does f(3+x)=f(3-x) mean, it means that no matter what value x takes, for example, x1, 3+x1 and 3-x1 are equal to the value of the function, which means that the image of y=f(x) is symmetric with respect to x=3.

You can see this by drawing this This is the method of graphics, I uploaded the picture for half a day and it is not displayed, Baidu garbage.

In addition diverges down.

If the function satisfies f(a+x)=f(a-x), then the function image is symmetrical with respect to x=aIf the function satisfies f(a+x)=-f(a-x), then the image is symmetrical with respect to (a,0) Supplement: For the sake of my attention, give it to me.

The answer should be d

f（x）=ax²+bx+c

Its axis of symmetry is the straight line x=-2a b

For the solution of the equation m[f(x)] nf(x)+p=0, let's take them as y1, y2

Then there must be y1=ax +bx+c, y2=ax +bx+c

Then from the image, y=y1, y=y2 is a straight line parallel to the x-axis.

They have an intersection point with f(x).

Due to the symmetry, the two solutions of the equation y1=ax +bx+c x1,x2 are symmetrical with respect to the straight line x=-2a b.

That is, 2(x1+x2)=-2a b

In the same way, the two solutions of the equation y2=ax +bx+c x3 and x4 should also be symmetrical with respect to the straight line x=-2a b.

Then we get 2(x3+x4)=-2a b

In answer c, we can find the axis of symmetry line x=, i.e. 1,4 is the solution of an equation and 2,3 is the solution of an equation.

So the set of solutions obtained can be.

In answer d, we can't find the axis of symmetry, which means that no matter how we group them, we can't make the sum of two equal to the sum of the other two.

Therefore the answer d is no.

What kind of question did you write, I can't see clearly, how can I help you do it?

Analysis: f(1)=0, substituting 1+2a+b=0 f(x)+1=0 with real root discriminant =4a -4(b+1) 0 (replacing a=-1- b 2 with a in ) b -2b-3 0

m^2+2am-2a=0 ..1(m-4)^2+2a(m-4)-1-2a=y...Formula 22-1, got.

15-8m-8a

Another 00 Da Dan teacher floated by.

Shenyang Zhimeng Education.

f(1)=0

f(1)=1+2a+b=0

a=-(1+b)/2

f(x)+1=x 2+2ax+b+1=0 has a solid root, then:

4a^2-4(b+1)>=0 (a)

(1+b)^2-4(b+1)>=0

(b+1)(b-3)>=0

b<=-1 (b<1) (i)

A<1

-(1+b)/2<1

b>-3 (ii)

i)(ii)=>-3-1<=a<1

Substitute (b) for (a) = >

4a^2+2a>=0

a>=0 (a>=-1, so rounding a<=-2)(2)m is the root of x 2+2ax+(b+1)=0, then m 2+2am+b+1=0

m^2+2am-2a=0

(m+a)^2+(a^-2a)=0

m+a)^2=(2a-a^2)=1-(a-1)^2<1=>-1-216-8-8=0

Therefore f(m-4) >0

<> is like a perfect brother, you look at the search for the world's reputation.

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