How to solve the system of quadratic equations with three variables in the second compulsory course of mathematics in senior one

How to solve the system of quadratic equations with three variables in the second compulsory course of mathematics in senior one

It can only be done by factorization. The difficulty lies in that there is no general method. We can only consider the countermeasures according to the specific situation of different problems. The basic principle is to reduce the number of equations by grouping decomposition

The inequality problem of quadratic equation,
6x²+17X-3≥0
（3X-1）*（X-1）≥X（2X-3）-5
sit back and wait
Two questions, and your answer should be x ＞ something,

(x+3)(6x-1)≥0
Greater than 0 is outside two
X ≤ - 3 or X ≥ 1 / 6
（3X-1）*（X-1）≥X（2X-3）-5
x²-x+6≥0
(x+2)(x-3)≥0
x≤-2,x≥3

Absolute value inequality and quadratic equation
It is known that a and B are real numbers, | a | + | B|

This (M + 1) (n + 1) > 0 and (M + 1) + (n + 1) > 0 have been proved, from which we can get (M + 1) > 0 and (n + 1) > 0, namely m > - 1, n > - 1
Similarly, M can be proved

Given a (1. - 2) B (2,4) C (4. - 3) d (x, 1) if the vector AB and the vector CD are collinear, what is the modulus of the vector BD

AB={1,6}
CD={X-4,4}
If the vector AB and the vector CD are collinear, then
1:X-4=6:4
So 6x-24 = 4
So x = 28 / 6 = 14 / 3
BD={X-2,1-4}={8/3,-3}
So | BD | = under root (64 / 9 + 9) = under root (145 / 9) = [under root (145)] / 3

D is a point on the extension line of BC, and the bisector of angle ABC and angle ACD intersects at point E

Angle a + angle ABC = angle ACD
Angle e + angle EBC = angle ECD
Angle EBC = 1 / 2 angle ABC
Angle ECD = 1 / 2 angle ACD
Angle a = 1 / 2 angle e

It is known that the midpoint of the ellipse C is at the origin, the focus is on the x-axis, the left and right focus are F1F2, and F1F2 = 2, point (1,3 / 2)
If the line L passing through F1 intersects with ellipse C at two points a and B, and the area of △ af2b is 12 root sign 2 / 7, the equation of line L is obtained

If P (1,3 / 2) of point F1 (- 1,0) and F2 (1,0) is on the ellipse, then 2A = Pf1 + PF2 = radical (4 + 9 / 4) + radical (9 / 4) = 5 / 2 + 3 / 2 = 4, that is, a = 2, C = 1, B ^ 2 = 4-1 = 3, that is, the elliptic equation is x ^ 2 / 4 + y ^ 2 / 3 = 1. Let the linear equation be x = ky-1

In the triangle ABC, the lengths of the opposite sides of the angles a, B and C are a, B and C respectively. Let a, B and C satisfy the condition that the square of B + the square of C - BC = the square of A,
In triangle ABC, the lengths of the opposite sides of angle a, angle B and angle c are a, B and C respectively. Let a, B and C satisfy the condition that the square of B + the square of C - BC = the square of a, and C / b = 1 / 2 + radical 3, then the values of angle A and tanb are obtained
The point is to find tanb

1. B ^ 2 + C ^ 2-bc = a ^ 2 shift B ^ 2 + C ^ 2-A ^ 2 = BC (b ^ 2 + C ^ 2-A ^ 2) / 2BC = 1 / 2 (b ^ 2 + C ^ 2-A ^ 2) / 2BC = cos60 so ∠ a = 60C / b = 1 / 2 + √ 3C / b = (1 + 2 √ 3) / 2 let C = (1 + 2 √ 3) XB = 2x (b ^ 2 + C ^ 2-A ^ 2) / 2BC = 1 / 2 solution a = √ 150.5 √ 3 / √ 15 = SINB / 2sinb = √ 5 / 5cosb = 2 √ 5 / 5

Given that the function f (x) = (x2 + KX + k) ex x is the value range of monotone decreasing K on the square (0,1), the monotone interval of the function can be obtained

1. When f (x) = (x ^ 2 + KX + k) e ^ XF '(x) = (2x + k) e ^ x + (x ^ 2 + KX + k) e ^ x = (x ^ 2 + (K + 2) + 2K) e ^ x = (x + 2) (x + k) e ^ Xe ^ x > 0 ∈ (0,1), f' (x) 0 can only be x + k = 0x > = - 2. In this case, the increasing interval of F (x) is [- 2, + ∞) and the decreasing interval is (- ∞, - 2). In this case, f '(x) > = 0x = - 2

To prove that the sum of the two right angles of a right triangle is equal to the sum of the diameter of its circumscribed circle and the diameter of its inscribed circle

Let the right angle △ ABC, C be the right angle, and the tangent points of the inscribed circle with the center O and AB, AC, BC are e, F, g respectively, which connect OE, of, OG. According to the tangent theorem, we can get CF = CG, AE = AF, be = BG, because the diameter of the circumscribed circle is equal to the hypotenuse: ab = AE + be = AF + BG, because AC = AF + CF, BC = BG + CG, cfog is a regular four edge

Given the set u = R, M = {x x ≥ 1}, n = {x x}

M ∩ n = φ (empty set)
∴CU（M∩N）=CU φ=U = R