On the image of inverse scale function y = K / X (K ≠ 0), there is a point a whose abscissa n makes the equation x & sup2;; - NX + n-1 = 0 have two equal real roots The area of the triangle enclosed by point a, B (0,0) and C (3,0) is equal to six. The analytic expression of the inverse scale function is obtained

On the image of inverse scale function y = K / X (K ≠ 0), there is a point a whose abscissa n makes the equation x & sup2;; - NX + n-1 = 0 have two equal real roots The area of the triangle enclosed by point a, B (0,0) and C (3,0) is equal to six. The analytic expression of the inverse scale function is obtained

1. Because X & # 178; - NX + n-1 = 0 has two equal real roots, so - (n & # 178;) - 4 * 1 * (n-1) = 0; so n = 2 (this step refers to Veda's theorem). Now we know that the X coordinate of point a is 2, which is temporarily set as a (2, y) 2. Because the area of the triangle formed by point a, point B (0,0) and point C (3,0) is equal to six, so from point a to X

It is known that the two points whose roots of the equation X-2 (R-3) x + k-4k-1 = 0 are abscissa and ordinate are just on the image of the inverse scale function y = m / X,
It is known that the two points whose roots of the equation X-2 (R-3) x + k-4k-1 = 0 are abscissa and ordinate are just on the image of the inverse scale function y = m / x, and M satisfying the condition is the minimum

If the discriminant = 4 (K-3) ^ 2-4 * 1 * (4K ^ 2-4k-1) > 0, K can be obtained

Point (1___ ）On the image of the function y = 5x-4, so x = 1, y=___ Is a set of solutions of the equation 5x-y = 4

Point (1,1) is on the image of function y = 5x-4, so x = 1, y = 1 are a set of solutions of equation 5x-y = 4

Why must the orbit equation of the center of the moving circle x ^ 2 + y ^ 2 - (4m + 2) x-2my + 4m ^ 2 + 4m + 1 = 0 be y = 1 / 2x-1 / 2 (Y > 0)?

Reduced to (x - (2m + 1)) ^ 2 + (y-m) ^ 2 = m ^ 2
The center (x0, Y0) satisfies x0 = 2m + 1, Y0 = M
The radius of the circle is required to be | m | 0
Y0 is not equal to 0, as if it does not have to be > 0