If the two roots of the quadratic equation x & # 178; + MX + 3-m = 0, one is greater than 2 and the other is less than 2, then the value range of M is

If the two roots of the quadratic equation x & # 178; + MX + 3-m = 0, one is greater than 2 and the other is less than 2, then the value range of M is

One ∵ has two intersections ∵ ⊿ ≥ 0, i.e. m-square-4ac ≥ 0 -------- (3)
Let two roots be a and B, and find a and B (denoted by M) a > B
∴a＞2---------（1）
b＜2--------（2）
The system of inequalities with simultaneous solutions (1) (2) (3)
The result is the range of M
If you make a graph, the symmetry line is - M / 2, (0,3-m) is certain, and then it will be clear at a glance.
Let f (x) = x ^ 2 + MX + 3-m
X-2y / x + y minus X-Y / x + y plus - x-4y / x + 4Y
The addition and subtraction of fractions in the first semester of grade two
Mathematical answer group for you to answer, I hope to help you
X-2y / x + y minus X-Y / x + y plus - x-4y / x + 4Y
=[x-2y - (X-Y)] / (x + y) plus - (x + 4Y) / (x + 4Y)
=[x-2y-x+y]/(x+y) - (x+4y)/(x+4y)
= -y/(x+y) - 1
（x-2y/x+y）-（x-y/x+y）+（-x-4y/x+4y）=（x-2y-x+y）/（x+y）-（x+4y）/（x+4y）=-y/x+y-1=-x-2y/x+y
If f (x) = | (x) = | (x) = | (x) = | (x) = | (x) = | (x) = | (x) = | (x) = | (x) = | (x) = | (x) = | (x) = | (x) = | (x) = |
When a = 0, x > = 0, f (x) = x, f (x) is an odd function on R, ∩ x = f (x)
When a ≠ 0, f (x) = {x-2a ^ 2, x > = a ^ 2;
{-x,0
Reading materials: if X1 and X2 are two of the univariate quadratic equations AX2 + BX + C = 0, then there are X1 + x2 = − Ba, x1x2 = ca. this is the relationship between the root and coefficient of the univariate quadratic equation, and we can use it to solve problems. We know that X1 and X2 are two of the equations x2-4x + 2 = 0, and find: (1) the value of 1x1 + 1x2; (2) the value of (x1-x2) 2
∵x1+x2=4，x1x2=2．（1） 1x1+1x2＝x1+x2x1x2＝42＝2；（2）（x1-x2）2=（x1+x2）2-4x1x2=42-4×2=8．
If the line 3x + 4Y + 2 = 0 and the circle X & sup2; + Y & sup2; + 4x = 0 intersect at two points a and B, then the equation of the vertical bisector obtained from the line AB is?
The center of X & sup2; + Y & sup2; + 4x = 0 is (- 2,0)
So the equation of the vertical bisector of the line AB is the vertical bisector of 3x + 4Y + 2 = 0 through the center of the circle
The equation is 4x-3y + 8 = 0
3 / 5-4y and 2Y plus 9 / 5 are opposite numbers, so what is y
The sum of 2Y and 4Y is 5-6
(3 / 5-4y) + (2Y + 9 / 5) = 0
3 / 5-4y + 2Y + 9 / 5 = 0
4y-2y = 3 / 5 + 9 / 5
2Y = 12 / 5
Y = 6 / 5
If f (x) = 2x-a / X (a is a real number) is defined as (0,1] (a is a real number) 1, if f (x) > 5 is constant in the domain, the value of a is obtained
It is known that the definition field of function f (x) = 2x-a / X (a is a real number) is (0,1] (a is a real number)
1. If f (x) > 5 is constant in the domain of definition, the value range of a is obtained
2. If the function y = f (x) is a decreasing function in the domain of definition, the value range of a is obtained
(1) ∵ f (x) = 2x-a / x > 5 holds in the domain of definition
∴2x*2-a>5x
2x*2-5x-a>0
Then Δ = (- 5) 2-4 × 2 × (- a)
two million four hundred and twenty-four thousand four hundred and fifty-five
If the two roots of equation 2x ^ 2 + 6x-3 = 0 are X1 and X2, then (x1-x2) ^ 2=_______ ,(X1+3)(X2+3)=_____
The two roots of 2x ^ 2 + 6x-3 = 0 are X1 and x2
∴x1+x2=-3,x1·x2=-3/2
(x1-x2)²
=(x1+x2)²-4x1·x2
=9+3/2×4
=15
(x1+3)(x2+3)
=x1·x2+3(x1+x2)+9
=-3/2-9+9
=-3/2
X & sup2; - 9 / 4Y & sup2; = 0 4 / 3x & sup2; + 4xy + 3Y & sup2; = 1 / 3 how to calculate the second, give 20 points
4X & sup2; + 12xy + 9y & sup2; = 1, i.e. (2x + 3Y) & sup2; = 1 the second solution: first solve the first equation: (x-3 / 2Y) (x + 3 / 2) = 0 to get x = 3 / 2Y or x = - 3 / 2Y. When x = 3 / 2Y, substitute into the second equation to get y = ± 1 / 6. When x = - 3 / 2Y, substitute into the second equation to get y = no solution, so x = 3 / 4, y = 1 / 6 or
[2x / (radical 3) + (radical 3) * y] square = 1 / (radical 3)
From the first, we can get xy = 3 / 2 or - 3 / 2, and then we can bring in the second formula through classification discussion. X = ± 3 / 2Y ????????????????????????????????????????????????????????????????????????????????????????? ？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？ ... unfold
From the first, we get xy = 3 / 2 or - 3 / 2, and then discuss the classification Just bring in the second formula. Question: x = ± 3 / 2Y ???????????????????????????????????????????????????????????????????????????????????????????????? ？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？
If the square of 2Y plus 3Y plus 7 equals 8, what is the square of 4Y plus 6y minus 9
fast
2Y2 + 3Y + 7 = 8, then 2Y2 + 3Y = 1,
Then 4y2 + 6y = 2 (2Y2 + 3Y) = 2,
So 4y2 + 6y-9 = 2-9 = - 7
From the square of 2Y plus 3Y plus 7 equals 8: the square of 2Y plus 3Y equals 1, so the square of 4Y plus 6y equals 2, so the square of 4Y plus 6y minus 9 equals - 7
The square of 2Y + 3Y + 7 = 8
It is reduced to the square of 2Y + 3Y = 1
And then we put * 2 on both sides of this equation
Becomes the square of 4Y + 6y = 2
What we want is the square of 4Y + 6y-9 =?
That's to square 4Y + 6y to 2
2-9=-8
So the square of 4Y plus 6y minus 9 = - 7