What is the maximum distance between circle x2 + y2-2x + 4Y + 4 = 0 and straight line 3x-4y + 9 = 0?

What is the maximum distance between circle x2 + y2-2x + 4Y + 4 = 0 and straight line 3x-4y + 9 = 0?

Maximum distance from circle to line = distance from center of circle to line + radius
(x-1)^2+(y+2)^2=1
Center (1, - 2), radius 1
So the maximum distance = | 3 + 8 + 9 | / 5 + 1 = 4 + 1 = 5

Let a be the moving point on the circle x ^ 2 + y ^ 2 = 1, then the maximum distance from a to the line 3x + 4y-10 = 0 is?

The maximum distance from the moving point on the circle x ^ 2 + y ^ 2 = 1 to the straight line 3x + 4y-10 = 0 is required
This distance = radius of circle + distance from center of circle to line
d=1+|-10|/5=3

The maximum distance from the upper moving point P to the line 3x + 4y-25 = 0 is

The distance from point to line is the vertical distance from point to line
So only the distance from the center of the circle to the straight line + half level is the maximum distance
Because the center of the circle (0,0), according to the distance from the point to the line, the formula is: D = 5
Dmax = 5 + 3 = 8

Known inequality (a + 1) x2 + ax + a > b (x2 + X + 1) holds for any real number, try to compare the size of real numbers a and B

(a+1)x2+ax+a>b(x2+x+1)
ax2+x2+ax+a>b(x2+x+1)
a(x2+x+1)+x2>b(x2+x+1)
a(x2+x+1)+x2-b(x2+x+1)>0
(a-b)(x2+x+1)+x2>0
If the inequality holds for any real number, then a must be greater than B

If the inequality 0 ≤ x2 + ax + 5 ≤ 4 about X has exactly one solution, then the real number a=______ ．

If the minimum value is less than 4, then x2 + ax + 5 ＜ 4 has more than one solution. If the minimum value is greater than 4, then there is no solution, so the minimum value = 4 ∧ 20 − A24 = 4 ∧ a = ± 2, so the answer is ± 2

The solution of inequality x2 + (m-1) x-m > 0 about X

Factorization into
(x+m)(x-1)>0
May be
x>-m x>1
or
If x0 and | m | > 1, then x > M
If M = 0, then x > 1
If M1, then 1 > x > M
If M

Solving inequality (3x2 + 2x-2) / (3 + 2x-x2)

(3x2+2x-2)/(3+2x-x2)

We know the inequality x2-x + 1 > 2x + M. (1) the solution of X inequality (2) if the inequality is constant on [- 1,1], find the value range of real number M
We know the inequality x2-x + 1 ＞ 2x + M. (1) the solution of X inequality (2) if the inequality holds on [- 1,1], find the value range of real number M

X2-x + 1 ＞ 2x mx2-3x ＞ M-1 let y = x2-3x, rigid y be symmetric with respect to x = 3 / 2, and pass through point (0,0) and point (3,0). In the interval [negative 1,1], the function is monotone decreasing, so when x = 1, y is the minimum, y = 1-3 = negative 2. If the equation holds, then negative 2 ＞ M-1, rigid, m ＜ negative 1

Finding the solution set of inequality 3 + 2x-x2 ≤ 0

X-2x-3 ≥ 0 (x-3) (x + 1) ≥ 0, X ≥ 3 or X ≤ - 1

What are the solution sets of inequality x ^ 2-2x-3 ＞ 0 and inequality x ^ 2-2x-3 ＜ 0?

x^2-2x-3＞0
(x-3) (x + 1) > 0, = > x > 3 and x > - 1 or X