Find the range of y x 2 2x 3 2 x 2 2x 1

2x^2-4x-1=2x^2-4x+2-3=2(x-2)^2-3>=-3

For convenience. Let a=2x 2-4x-1>=-3y=4 a, obviously a is not equal to 0

If -3<=a<0,0<-a<=3, then 1 (-a)>=1 3, 1 a<=-1 3, so 4 a<=-4 3

If a>0, then 4 a>0

So the range (- 4 3] (0,+.)

Define the domain 5x-2>=0, x>=2 5, and x is not equal to 0, so x>=2 5

Let a = (5x-2).

then x=(a 2+2) 5

So y=a x=5a (a 2+2)=5 (a+2 a) this is the checkmark function, a>0.

0 2 is an increment function.

Now a>=2 5

So a = 2, a + 2 a minimum = 2 2

So a+2 a>=2 2

0<1/(a+2/a)<=1/2√2=√2/40<5/(a+2/a)<=5√2/4

So the range (0,5 2,4].

Let a = (2x-3).

x=(a^2+3)/2

y=4(a^2+3)/2-1+a

2a^2+a+5

2(a+1/4)^2+39/8

a=√(2x-3)>=0

So the definition domain is on the right side of the axis of symmetry a=-1 4.

The opening is upward. So it's an increment function.

a=0, ymin=5

So the value range [5,+

y=x^2-4x+3/2x^2-x-1

x-1)(x-3) (x-1)(2x+1)(x-3) Rock Calendar (2x+1) x is not 1

1/2-7/2*(2x+1)

1 2-7 key celery (4x+2).

x is not 1 2

Therefore, the only values that can not be obtained by y are: rough search 1 2 and 2 3, so the value range is not 1 2 and 2 3 for the whole set of real numbers!

y=(2x 2+4x-7) (x 2+2x+3)x 2y+2yx+3y=2x 2+4x-7y-2)x 2+2(y-2)x+3y+7=0, this square thought type about x has a solution, then the discriminant Yutong is greater than or equal to 0, so 4(y-2) 2-4(y-2)(3y+7)>=0y-2)(y-2-3y-7)>=0

y-2)(2y+9)<=0

9 2<=y《Guess the Split=2

So the value range [-9 2,2].

Easy to obtain: Define the domain as r

Deformation y(x +x+1) = x -2x-3

y-1)x²+(y+2)x+y+3=0

Think of this equation as an equation about x:

1) When y=1, 3x+4=0, get: x=-4 3, so y=1 is preferable;

2) When y≠1, 0 4y +8y-12(y+2) -4(y-1)(y+3) 0

3y²-4y+16≧0

3y²+4y-16≦0

gets: (-2-2 13) 3 (-2+2 13) 3, and y≠1 is summed up, (-2-2 13) 3 (-2+2 13) 3, which is the range

Have fun! I hope it can help you, if you don't understand, please ask, I wish you progress! o(∩_o

y=(x²-2x-3)/(x²+x+1)

yx²+yx+y=x²-2x-3

y-1)x²+(y+2)x+y+3=0

When x=-4 3, y=1

When x≠-4 3, y≠1, (y-1)x +(y+2)x+y+3=0 is a quadratic equation about x, and this equation has a real solution, so there is.

(y+2) -4(y-1)(y+3)>=0, i.e., -3y-4y+16>=0

2-2 13) 3<=y<=(-2+2 13) 3

The range of y=(x -2x-3) (x +x+1) is (-2-2 13) 3<=y<=(-2+2 13) 3

y(x^2+1)=x^2+2x+3

1-y)x^2+2x+3-y=0

When y=1, x=-1 holds.

When y<>1, the quadratic equation for x has a real root, and the discriminant equation = 0, that is, 4-4*(1-y)(3-y)>=0

1-(3-4y+y^2)>=0

2+4y-y^2>=0

y^2-4y+2<=0

2-Root2<=y<=2+Root2

y=0 is also true.

Therefore, the value range is [2-root2, 2+root2].

y=[2(x^2+x)+3]/(x^2+x)=2(x^2+x)/(x^2+x)+3/(x^2+x)=2+3/(x^2+x)

x 2+x=(x+1 2) 2-1 4>=-1 4, so if -1 4<=x 2+x<0

then 1 (x 2+x)<=-4

If x 2+x>0

then 1 (x 2+x) > 0

So 1 (x 2+x)<=-4,1 (x 2+x)>02+1 (x 2+x)<=-2,2+1 (x 2+x)>2 so the range (- 2] (2,+

Solution: Formula the function formula to obtain: y=-x2-2x+3=-(x+1)2+4, the definition domain of the function is r, so the maximum value of the function is 4, so that the value range of the function is: (-4].

So the answer is: (-4].