It is known that the square of the equation KX of x minus (3K-1) x + 2 (k-1) = 0. Proof: no matter what the real number k is, the equation always has real number It is known that the square of the equation KX minus (3K-1) x + 2 (k-1) = 0 Proof: no matter what the real number k is, the equation always has real roots (2) If this equation has two real roots, x1, X2, and | x1, X2 | = 2, find the value of K?

It is known that the square of the equation KX of x minus (3K-1) x + 2 (k-1) = 0. Proof: no matter what the real number k is, the equation always has real number It is known that the square of the equation KX minus (3K-1) x + 2 (k-1) = 0 Proof: no matter what the real number k is, the equation always has real roots (2) If this equation has two real roots, x1, X2, and | x1, X2 | = 2, find the value of K?

It is known that the square of the equation KX about X is one (3K-1) x two (k-1-2 O. no matter what the real number of K is, the equation always has real root
If the equation has two real roots x1, X2, and x1-x2-22, find the value of K

K equals three

If there are two intersections between the curve y = x-square-x + 2 and the straight line y = x + m, what is the value range of the real number m

Substituting y = x + m into y = x2-x + 2, 0 = x2-2x-m + 2, that is, (x-1) 2-m + 1 = 0, because (x-1) 2 is greater than or equal to 0, - M + 1 must be less than or equal to 0, so m is greater than or equal to 0

Find the trajectory equation of the point (x + y, XY) when the point moves on the circle with the origin as the center and a as the radius according to (x, y)

The equation of circle: x ^ 2 + y ^ 2 = a ^ 2
From the parametric equation of circle: x + y = a * cos θ, xy = a * sin θ
Because cos θ ^ 2 + sin θ ^ 2 = 1
So the trajectory equation: (x + y) ^ 2 + x ^ 2Y ^ 2 = a ^ 2
Glad to solve the problem for you!