![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
It is known that the real numbers a and B are two of the equation x + m-2 + 1 = 0 about X. find the value of (1 + Ma + a) (1 + MB + b)
According to Weida's theorem, a + B = 2-m, ab = 1
The result is: m (a + b) ^ 2 + (a + b) ^ 2-AB + AB (m) ^ 2 + 1
That is: m (2-m) ^ 2 + (2-m) ^ 2 + m ^ 2
The result is: m ^ 3-2 (m) ^ 2 + 4
Given that a and B are two of the equations x2 + (m-2) x + 1 = 0 about X, then the value of (1 + Ma + A2) (1 + MB + B2) is______ ．
∵ A and B are two of the equations x2 + (m-2) x + 1 = 0, ∵ A2 + (m-2) a + 1 = 0, B2 + (m-2) B + 1 = 0, ab = 1, ∵ 1 + Ma + A2 = 2A, 1 + MB + B2 = 2B, ∵ (1 + Ma + A2) (1 + MB + B2) = 2A, 2b = 4AB = 4
If the two real roots of the quadratic equation x ^ 2 + (m-1) x + 2004 = 0 are a and B, then (a ^ 2 + Ma + 2004) (b ^ 2 + MB + 2004) =?
X ^ 2 is the square of X
Because a and B are two equations,
So there are:
a^2+(m-1)a+2004=0
b^2+(m-1)b+2004=0
After deformation, a ^ 2 + Ma + 2004 = a
b^2+mb+2004=b
So (a ^ 2 + Ma + 2004) (b ^ 2 + MB + 2004) = ab
According to the relationship between root and coefficient, ab = 2004
We know two real root solutions a, B of the quadratic equation x ^ 2 + (m-5) x + 9 = 0 with respect to X. (1) calculate (a ^ 2 + Ma + 9) (b)
^2 + MB + 9) (2) find the minimum value of a ^ 2 + B ^ 2
X^2+(M-5)X+9=0
a+b=5-m,ab=9
x^2+mx+9=5x
(A^2+MA+9)(B^2+MB+9)
=5a*5b=25*ab=25*9=225
2)
A^2+B^2=(a+b)^2-2ab=(5-m)^2-18=7-10m+m^2=(m-5)^2-18
When m = 5, the minimum value is - 18
-------These are two questions. We should add points!
RELATED INFORMATIONS
- 1. Given that a and B are two of the equations x2 + (M + 2) x + 1 = 0, then the value of (A2 + Ma + 1) (B2 + MB + 1) is______ ．
- 2. Quadratic factorization factor of 4a-12ab + 9b-c Come on, come on
- 3. Given that 14a2 + 9b2 − a + 12b + 5 = 0, find the value of (a − 2) 2A2 − B2
- 4. Let a = {X / (x + 3) (2-x)} > = 0, B = {X / x ^ - 4ax - (1-4a ^)
- 5. Factoring the square of B - the square of 4A 2A (m-n) - 4B (n-m) remember
- 6. If 0.3A ^ m + 1b ^ 5 is the same as 4A ^ 2B ^ n-1, then M=_ ,n=_?
- 7. Finding the midpoint trajectory equation of parallel string with slope 2 in ellipse x ^ 2 + 2Y ^ 2 = 1 Let the linear equation l with slope 2 be y = KX + B = 2x + B Simultaneous: x ^ 2 + 2Y ^ 2 = 1, y = 2x + B 9x＾+8bx+2b＾-1=0 L intersects the ellipse at two points a (x1, Y1) and B (X2, Y2) x1+x2=-8b/9=2x x=-4b/9 b=-9x/4 y1+y2=2（x1+x2）+2b=4x+2b=2y y=2x+b=8x/4-9x/4=-x/4 ∴4y+x=0 Why is x = 8x / 4 in the last y = 2x + B = 8x / 4-9x / 4 = - X / 4 not before x = - 4B / 9 How did this step come about?
- 8. Find the trajectory equation of the midpoint m of a group of chords with slope 2, where the square of 8 / x plus the square of 4 / y of the ellipse equals 1
- 9. The trajectory equation of the midpoint of a group of parallel strings with slope k of ellipse x2 + Y2 / 4 = 1 is a part of the straight line 2x + y = O, and K is obtained
- 10. Given that circle B: (x + 1) ^ 2 + y ^ 2 = 16 and point a (1,0), C is any point on the circle, the trajectory equation of the intersection P of AC vertical bisector L and segment CB is obtained It is better to set P (x, y) as any point on the trajectory
- 11. Given the points (2,8) and B (6, - 4), find a point m on the x-axis such that Ma = MB, and find the coordinates of point M
- 12. In this paper, we use the distribution law m (a + B + C) = ma + MB + MC to calculate. Given y ^ 2 + y + 1 = 35 / 6, a + B + C = 36 / 7, we can find a (y ^ 2 + y + 1) +The value of B (y ^ 2 + y + 1) + C (y ^ 2 + y + 1) Speed, online, etc
- 13. As shown in the figure, two objects a and B are stacked on a smooth horizontal plane. It is known that Ma = 6kg, MB = 2kg, and the dynamic friction factor between a and B is μ = 0.2. A thin line is tied on object a, and the maximum tensile force that the thin line can bear is 20n. Now the thin line is pulled horizontally to the right, g = 10m / S2 Find (1) if the tension is 20n, then the friction between ab? (2) In order to make AB relatively static, the value range of horizontal tension f is calculated
- 14. The mass of the two objects is Ma = 2kg, MB = 4kg, the dynamic friction coefficient between a and B is u = 0.2, and the dynamic friction coefficient between B and the ground is u = 0.4, A. The relative sliding occurs between B and the ground, and the friction between the contact surfaces is calculated
- 15. It is known that Ma = MB = m, all friction forces can be ignored, and the mass of rope and pulley can be ignored. When a and B are released from rest, and B drops by 3 / 4m, B will lose weight
- 16. If the angle xoy is 60 ° and M is a point in the angle xoy, and the distance from it to ox is Ma = 2, and the distance from it to oy is MB = 11, then OM=_____ I hope it's quick and the process is detailed
- 17. Two high school mathematical inequality questions 1. The known real number A.B.C satisfies a + B + C + D = 3. A ^ + 2B ^ + 3C ^ + 6D ^ = 5 What are the maximum and minimum values of a? 2. If a and B belong to R +, and ab - (a + b) = 1, what is the minimum value of a + B? Thank you very much
- 18. Ask about inequality, A3 + B3 + C3 + 3ABC > 2 (a + b) C2 a > 0, b > 0, C > 0, a + b > C, a, B, C are not equal to each other. Who can help me to prove that? Where A3, B3, C3 is the cube of a, B, C, C2 is the square of C,
- 19. Compare the values of 3x2 + 4 and 2x2 + 4x by collocation method
- 20. When a < 5 is known, the solution set of the inequality ax ≥ 5x + A + 1 is______ ．