Ask a master of solving a math problem in the evaluation range!!

domain: defines the domain of all real numbers except -1, 3, and 1(Of course, when x takes -1 3 and 1, the denominator is 0, which is meaningless.) ）

range: value range.

Divide the numerator and denominator on the right side of the equation by x 2 at the same time, and we get h(x)=6 (3-2 x-1 (x 2));

Let g(x)=1 h(x), i.e., g(x)*h(x)=1.

then g(x)=(3-2 x-1 (x 2)) 6=1 2-1 (3x)-1 (6x 2)

Let a=1 x, then g(x)=1 2-1 3*a-1 6*a 2, using the parabola of the quadratic function, throw away a=-3 and 1, and obtain the value range of g(x) as (-infinity, 0)u(0,2 3).

So the range of h(x)=1 g(x) is (-infinity,0)u[3 2,+infinity).

Is there a problem with the range in the question? You can get 3 2, when x=-1, h(x)=3 2.

Or is range not the meaning of a range?

In short, I think the answer is (-infinity,0)u[3 2,+infinity).

What does it mean in English? -

h(x)=6 (3-2(1 x)-(1 x 2)) let t=1 x

The value range of 3-2t-t 2 can be calculated.

Therefore, the range of values of 6 (3-2t-t 2), i.e., the value range, can be found.

3x^-2x-1≠0

x≠2 and -1 3

Define domain natural.

You can use the discriminant method.

y=6x^/(3x^-2x-1)

3y-6)x^-2yx-y=0

4y^+12y^-24y≥0

y≥3/2 y≤0

The range is (-infinity,0)u[3 2,+infinity).

The answer is questionable, the definition range is the same, the value range should be h 3 2 or h 0 validation: when x -1 gives h 3 2, it contradicts the h you give is greater than 2.

Method: You can use the exchange method given upstairs.

It is also possible to put the equation, with x as the unknown number, and h in the coefficient

Then let b 2-4ac 0, and the range of h can be solved.

The discriminant formula is simpler, organized as (y-2)x 2+(y+1)x+(y-2)=0, the denominator is evergrandly higher than 0, and the defined domain x belongs to the real number r, then the discriminant formula =(y+1) 2-4(y-2) 2 is greater than or equal to zero, that is, y 2-6y 5 is less than or equal to zero, and 1 is less than or equal to x less than or equal to 5

Method 1:

Match the molecule to x-2+3

then y=-1+3 (2-x).

Because 2-x 2 and ≠ 0

So 1 (2-x) 1, 2, or 0

So 3 (2-x) 3, 2, or 0

So y=-1+3 (2-x) 1 2 or -1 is in the range (- 1) [1, 2,+

Method 2: 2y-yx = x +1

x²=(2y-1)/(y+1)

Because x 0

So (2y-1) (y+1) 0

Solving this inequality yields y (-1) [1 2,+

y=(x 2+1) (2-x 2)=(3+x 2-2) (2-x 2)=3 (2-x 2)-1, that is, find the range of 2-x 2, 2-x 2"2, and not equal to 0, when 0<2-x 2"2, 1 2"y"positive infinity; When 2-x 2<0, negative infinity

y=(x²+1）/(2-x²)=( x²-2+3)/(2-x²)=-1 + 3/(2-x²)

Because: 2-x 2, the value range of 3 (2-x) is: (-infinite to 0) (0 to 3 2).

So the total range is (minus infinity to -1) (-1 to 1 2).

Find the domain of the function y=x -ax+1,x [-1,1].

f(x)=x²-ax+1=(x-a/2)²+1-a²/41)a∈(-2]:

min[f(x)]=f(-1)=2+a

max[f(x)]=f(1)=2-a

2)a∈(-2,0)

min[f(x)]=f(a2)=1-a and megamu 4max[f(x)]=f(1)=2-a

3)a∈[0,2)

min[f(x)]=f(a 2)=1-a Guess 4max[f(x)]=f(-1)=2+a

4) a [2,+ calling).

min[f(x)]=f(1)=2-a

max[f(x)]=f(-1)=2+a

a|≥2:min[f(x)]=2-|a|

a|<2:min[f(x)]=1-a²/4max[f(x)]=2+|a|

x>1;Let t=x-1>0;So y=x+4 (x-1)=(x-1)+1+4 (x-1)=t+4 t+1;

So when t=2, y takes the minimum value of 5;

So y>=5

y=(x 2+2x+1) x (x 0, divide by x)=x+2+(1 x).

x+(1/x)+2

x＞0,x+(1/x)≤2

y≤2+2=4

value range [4, positive infinity).

Note: y=x+(k x) (k 0).

Then y 2 * root number k

Divide x by in, put the x term together, because x is greater than zero, so the x term is greater than two, so its value range is greater than or equal to four.

The range of functions can be solved using the commutation method.

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