Proof: when m is a real number, at least one of the quadratic equation x2-5x + M = 0 and the equation 2x2 + x-6-m = 0 has a real root

Assuming that the above two equations have no real roots, then △ 1 = 25 − 4m ＜ 0, ① △ 2 = 1 + 4 × 2 × (6 + m) ＜ 0, ② ① get m ＞ 254, ② get m ＜ 498, such that m does not exist. At least one of the equations has a real root

It is known that the image of parabola y = 2x ^ 2-mx-2m has two intersections (x1,0) (x2,0) with X axis and X1 ^ 2 + x2 ^ 2 = 5, so we can find the intersection of M
X1 ^ 2 + x2 ^ 2 = 5 (square of X1 + square of x2 = 5)

Parabola y = 2x & sup2; - mx-2m. ⊿ = M & sup2; + 8m ≥ 0. = = = > m ≤ - 8, or m ≥ 0. According to Weida's theorem, X1 + x2 = m / 2, x1x2 = - M.. 5 = X1 & sup2; + x2 & sup2; = (x1 + x2) & sup2; - 2x1x2 = (M & sup2 / 4) + 2m. ∧ M & sup2; + 8m-20 = 0. = = > M1 = - 10, M2 = 2, ∧ M = - 10, or 2

It is known that the parabola y = x2-2x + m intersects the X axis at points a (x1, 0), B (X2, 0) (x2 & gt; x1), (1) if the point P (- 1, 2) is on the parabola y = x2-2x + m, find the value of M; (2) if the parabola y = AX2 + BX + m and the parabola y = x2-2x + m are symmetric about the Y axis, and the points Q1 (- 2, Q1) and Q2 (- 3, Q2) are on the parabola y = AX2 + BX + m, then the size relationship of Q1 and Q2 is___ (please write the conclusion on the horizontal line, do not write the answer process); (friendly tips: the conclusion should be filled in the corresponding position on the answer sheet) (3) let the vertex of the parabola y = x2-2x + m be m. if △ AMB is a right triangle, find the value of M

(1) ∵ the point P (- 1,2) is on the parabola y = x2-2x + m, (1 point) ∵ 2 = (- 1) 2-2 × (- 1) + m, (2 point) ∵ M = - 1. (3 point) (2) Q1 & lt; Q2 (7 point) (3) ∵ y = x2-2x + M = (x-1) 2 + M-1 ∵ m (1, m-1); (8 points) ∵ the parabola y = x2-2x + m has an upward opening and intersects with the X axis at points a (x1, 0), B (X2, 0) (x1 & lt; x2), ∵ M-1 & lt; 0, ∵ △ AMB is a right triangle, and am = MB, ∵ AMB = 90 ° △ AMB is an isosceles right triangle, (9 points) cross m as Mn ⊥ X axis, and the perpendicular foot is n. then n (1,0), NM = Na. ∵ 1-x1 = 1-m, ∵ X1 = m, (10 points) ∵ a (m, 0), ∵ m2-2m + M = 0, ∵ M = 0, or M = 1. (12 points)

It is known that the parabola y = x & sup2; - 2x + m intersects the X axis at points a (x1,0), B (x2,0) (x1 > x2)
(1) If the point P (- 1,2) is on the parabola y = x & sup2; - 2x + m, find the value of M
(2) If the parabola y = ax & sup2; - 2x + m is symmetric with respect to the y-axis, and the points Q1 (- 2, Q1), Q2 (- 3, Q2) are all on the parabola y = ax & sup2; + BX + m, then the size relation of Q1 and Q2 is () (please write the conclusion on the horizontal line, and there is no need to write the short answer process)
(3) Let the vertex of the parabola y = x & sup2; - 2x + m be m. if △ AMB is a right triangle, find the value of M

Let the same root be x = a, then a ^ 2-2a + M-3 = 0 (1) a ^ 2-3a + 2m = 0 (2) (1) - (2) get a-m-3 = 0, so a = m + 3, substituting a = m + 3 into (1) get (M + 3) ^ 2-2 (M + 3) + M-3 = 0, then M1 = 0, M2 = - 5, then A1 = 0 + 3 = 3, A2 = - 5 + 3 = - 2. So the value of M is 0 or - 5; when m = 0, the two formulas are the same

Given that circle C is symmetric about y axis, passing through point (1,0) and divided into two segments by X axis, the arc length ratio is 1:2, then the equation of circle C is ()
A. (x±33)2+y2＝43B. （x±33）2+y2=13C. x2+（y±33）2=43D. x2+（y±33）2=13

Let C (0, a) be the center of the circle, then the radius is ca. according to the ratio of the arc length of the circle divided into two segments by x-axis is 1:2, the center angle of the chord pair cut by x-axis is 2 π 3, so tan π 3 = | 1a |, a = ± 33, radius r = 43, so the equation of the circle is x2 + (Y ± 33) 2 = 43, so C

Given that circle C is symmetric about X axis, the center of circle C is on the straight line x-y-2 = 0, and circle C passes through the origin, the standard equation of circle C is obtained
If line L passes through point P (4,6) and is tangent to circle C, the equation of line L is obtained

If the circle C is symmetric about the X axis, the center of the circle is on the X axis. So let the center of the circle (x, 0) be brought into x-y-2 = 0 to get the center coordinates (2,0). Let the square of the equation (X-2) + the square of y = R cross the origin to satisfy the above equation. Let r = 2 be brought into the equation of the circle. Do you know
Let y = KX + B, that is, kx-y + B = 0
Through (4,6), we can get 6 = 4K + B
Draw a picture to see that the distance from the center of the circle to L is the radius
Therefore, according to the formula of distance from point to line, an equation can be obtained
The results can be obtained from the above formula

Given that circle C and circle (x-1) ^ 2 + (Y-2) ^ 2 = 1 are symmetric about X axis, then the equation of circle C is

∵ circle C and circle (x-1) ^ 2 + (Y-2) ^ 2 = 1 about X-axis symmetry, the center coordinates of known circle are (1,2)
The C coordinate of the center of the circle is (1, - 2), and the radius is the same as the known radius of the circle, which is 1
The equation of circle C is: (x-1) &# 178; + (y + 2) &# 178; = 1

Try to use the chord length t passing through the origin as the parameter to write the parametric equation of circle (x-a) ^ 2 + y ^ 2 = a ^ 2

Obviously, it is very obvious that it is very obvious that (x, y) on the circle (x, y) and the distance relationship of the origin (x, y) and the distance relationship of the origin distance from X (x, y) and the distance relationship of the origin (x \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\(2r-1) / 2R

The standard equation for a circle with a radius of 2 and a center at the origin is? Solved geometrically,

The distance between a point on a circle and the origin = the square of radius = 2 & # 178; = 4 = the square of abscissa + the square of ordinate = x & # 178; + Y & # 178; (Pythagorean theorem)
x²+y²=4

Write the parametric equation of radius 3 with the center of the circle at the origin of the coordinate
Urgent,

y=3sina
x=3cosa

Proof: when m is a real number, at least one of the quadratic equation x2-5x + M = 0 and the equation 2x2 + x-6-m = 0 has a real root