Urgently: Ask three high school math questions

1 -10 ≤3x-4 ≤5

2 ≤x ≤3

2 x 2 belongs to [1,4].

value range [2,8].

y=- x range (minus infinity, 0].

3 g(x)=f[f(x)]=[x^2+m]^2+mx^4+2mx^2+m^2+m

1.Remember y=3x-4, then the value of x=1 3(y+4)y is [-10,5], then the value of x is [-2,3], 1 x 4, so 2 8 x 8

x [0,+oo), root number x [0,+oo) so y=- root number x (-oo,0].

1.-10<3x-4<5

6<3x<9

20 root number x>0 - root number x<0 range (-infinity, 0]g(x)=(x 2+m) 2+m

x^4+m^2+2mx^2+m

x^4+2mx^2+m^2+m

1.From the meaning of the title, it can be seen that -10<3x-4<5

Solution -22(1) From the meaning of the title, it can be seen that 1<=x2<=4

Therefore, 2<=y<=8 means that the value range is [2,8].

2) From the meaning of the title, it can be seen that ye (negative infinity, 0).

3 g(x)=f(x2+m)=（x2+m)2+m=x4+2x2m+m2+m

1. g(x)=-x 2-3, f(x) is a quadratic function, and f(x)+g(x) is an odd function.

Therefore, the sum of g(x) and f(x) quadratic terms is 0, (-3) + c=0 (monotonicity and over origin).

So f(x)=x2+bx+3

When x [-1,2], the minimum value of f(x) is 1b = (2 * root number 2) (vertex method) b = 2 * root number 2 rounded off (interval maximum).

So f(x)=x2-2(root number2)x+3

1: Let f(x)=ax 2+bx+c, then f(x)+g(x)=(a-1)x 2+bx+c-3, because f(x)+g(x) is an odd function, so a=1, c=3, f(x)=x 2+bx+3, because when x [-1,2], the minimum value of f(x) is 1, so f(-1)>=1, f(2)>=1, the solution is -3<=b<=3, so the axis is x=-b 2 in the range of (, closed interval, the sign is not easy to play), discuss when in (-1, (at this time the minimum value of 1 is its lowest point), find b; When (,-1), the minimum value is f(-1)=1, find b.

2: f(x)=ax 2+bx+c, if a>b>c, and f(1)=0, so a+b+c=0, so b 2-4ac=(a+c) 2-4ac=(a-c) 2>0 (because a>c), then two solutions, i.e., two zeros.

3: log2f(a)=a, and because a≠1, so a=2, because f(log2a)=-b, so b=-2, so f(x)=x 2-x+2, the minimum value is 7 4, at this time, log2x=1 2, x=root number

To be honest, I really want to help you, but can't do anything about it. Actually, I'm very interested in mathematics, but because I haven't touched mathematics for 2 years, and I don't need to study mathematics in college, I'm completely unfamiliar, and I don't know what some meanings are, such as odd functions, I don't know what they are, and I failed!

1.Because a is known to be the angle of the second quadrant and sina = root number 15 2 so cosa = -1 4

So sin(a+4) = root number 2 2 times (sina + cosa).

Because 1 = the square of the sina + the square of the cosa sin2a = 2sinacosa cos2a = the square of the cosa - the square of the sina .

So sin2a+cos2a+1=2sinacosa+2cosa's square=2cosa(sina+cosa).

Compare the root number 2 by 2 divided by 2cosa = - root number 2

So the above is all divided.

Root number 3) sin12-3cos12 divided by cos12

Because 2cos squared 12'-1=cos24

So 4cos squared 12'-2=2cos24

Compare to the other.

Root number 3) sin12-3cos12 divided by 2sin12cos12cos24

The above formula extracts 2 roots and 3 roots.

2 (root number 3) multiplied by one-half sin12 minus half root number three, cos12 = 2 (root number 3), sin(12-60) = 2 (root number 3), sin(-48) = -2 (root number 3), sin48

Below you get one-half sin48

So the original value is -4 (root number 3).

3.Because cos(a+ 4)=3 5 is positive, and because 2 is less than a and less than 3 2, a+ 4 is greater than 3 2

So 5 4 is less than a less than 3 2 so 5 2 is less than 2a is less than 3

And because cos(a+ 4)=3 5, cos(2a+ 2)=2cos 2(a+ 4)-1=-7 25=-sin2a

So sin2a=7 25 cos2a=-24 25

So the cos(2a+4) formula is reversed and you can do the math yourself.

I haven't checked it, you can do the math yourself and see if it's right or wrong.

The value range is [0,+ y=f(x) and y=a have two intersections, and when a>1, y=f(x) and y=a have only one intersection.

So a (0,1).

0, the original inequality becomes x+2>x-2 so x>0x=0, the original inequality becomes x+2>1 so x=0x<0 the original unequal smooth becomes x+2>(x-2) -1x+2> liquid 1 (x-2) (x-2)(x+2)<1 x 2<5 So-root number 5 - root number 5

10. a<0,b<=0

The first: let f(x)=x 2+kx+2k-1....From the interval, it can be obtained: the two are unequal, so the discriminant formula "0", and then draw a rough sketch, f(-2)>0, f(-1)<0, f(1)<0, f(2)>0, and the range of k can be solved;

The second is to divide the two molecules into 1=a+(1-a), 4=4a+4(1-a), and then calculate them separately, using the basic inequality to make it;

The third: I don't know if your root number is **, -4a 2-b 2=4ab-(2a+b) 2=4ab-1, and then it's only related to ab, using basic inequalities!

(1) =16m 2-4*4(m+2)>=0 to obtain m>=2 or m<=-1 a +b =(a+b) -2ab

a+b=m ab=m+2 4 substituting m -1 2m-1 is obtained from the range of values of m, when m=-1, the minimum is there.

2) The axis of symmetry is obtained from f(5-x)=f(x-3) and is 5-x+x-3 2=1;Get b = -2a

f(x)=x has equal roots to get =0 b=1 a=-1 2

Note: Make the definition domain [m,n] and the value range [3m,3n]. is both an intersection problem with y=3x (?). y=-1 2x +x=3x x=0 x=-4 is m=0,n=-4;

3 For example, y=x you see that it fits or not.

Refuse to answer the questions with too many points and few points.

,ab=(m+2)/4

a 2 + b 2 = (a + b) 2-2ab = m 2-m 2-1 and when m = 1 4 there is a minimum value of -17 16.

i.e. a(-x+5) 2+b(-x+5)=a(x-3) 2+b(x-3).

ax^2-10ax+25a-bx+5b=ax^2-6ax+9a+bx-3b

4a+2b)x+(-16a-8b)=0

4a+2b=0 and -16a-8b=0

The solution gives a=-b 2

f(x)=x has equal roots, i.e., ax 2+bx=x. ax^2+(b-1)x=0

The discriminant formula is equal to zero i.e.

b-1)^2-4a*0=0

b-1=0b=1

a=-b/2=-1/2

f(x)=-x^2/2+x

f(m)=3m, and m=0 or -4

exists, defines the domain [-4,0], and the value range [-12,0].

3 False, if the odd function, then f(x) is still an increment function on (- 1).

1；a a+b=m a*b=[m+2 4] The answer to answer is a by using the recipe method

Let the straight angle edge be a, the other right angle side is b, and the hypotenuse is, when the circumference is equal, the area of the isosceles right triangle is the largest, that is, when the area is the largest when a=b, then there is c2=2a2, c=root number 2a, then there is: 2a + root number 2a = 1 + root number 2, and the solution is a = (1 + root number 2) (2 + root number 2).

then triangle area =

The axis of symmetry of the function is x=a

So as long as a<1 or a>2 is OK

Refuse to answer the questions with too many points and few points.

1, (-infinity, -2), (2, + infinity).

2、a3、m

Solution: 1, w=a+1 a (a≠ 1).

a>0, w=a+1 a>2*root number(a*1 a)=2a<0,w=-[a)+(1 a)].

w=-[a)+(1 a)]>2*root number(-a*-1 a)=2w<-2

So w<-2, w>2

2、ab-(a+b)=1

a-1) (b-1) = 2< = 2 a-1>0, b-1>0 (a-1<0 or b-1<0 (a-1) (b-1) = 2 is not true).

2< = 2 on both sides of the root number.

a+b>=2+2 root number 2 a=b=1 + root number 2 is equal sign.

3. x>0, y>0 and x is not equal to y

m-p=x/y+y/x>0 m>p

m-q=xy+1 (xy)-2> 2 root number [xy*1 (xy)]+2=0 m>q

m-n=y x+x y-2> 2 root number (x y*y x)+2=0 m>n

So m is the maximum.

x=2kπ+x

Get x=k ,k z

2. x1+x2=-3 3, x1x2=4, from which we can see that the two are negative.

tan（a+b）=（tana+tanb）/（1-tanatanb）=（x1+x2）/（1-x1x2）

Because a, b (- 2,0).

So a+b=-2 3

3. Certificate: cos [arcsin(1 a)+arcsin(1 b)] = cos90 degrees = 0

a 2-1) a* (b 2-1) b-1 a*1 b=0 multiply both sides by ab to get (a 2-1)* b 2-1)=1(a 2-1)*(b 2-1)=1

a^2b^2-(a^2+b^2)+1=1

Because in a right triangle, c 2 = a 2 + b 2

So c=ab, take the logarithm, lgc=lgab

So LGC=LGA+LGB