If point a is any point on the circle x2 + Y2 + ax + 4y-5 = 0 and the symmetric point of a on the line x + 2y-1 = 0 is also on the circle C, then the real number a is equal to______ .

If point a is any point on the circle x2 + Y2 + ax + 4y-5 = 0 and the symmetric point of a on the line x + 2y-1 = 0 is also on the circle C, then the real number a is equal to______ .

Point a is any point on the circle x2 + Y2 + ax + 4y-5 = 0, and the symmetric point of a about the straight line x + 2y-1 = 0 is also on the circle C, which means that the straight line passes through the center of the circle, and the coordinates of the center of the circle (- A2, − 2) are substituted into the linear equation x + 2y-1 = 0, so − A2 − 4 − 1 = 0, so a = - 10, so the answer is: - 10
If point a is any point on the circle x2 + Y2 + ax + 4y-5 = 0 and the symmetric point of a on the line x + 2y-1 = 0 is also on the circle C, then the real number a is equal to______ .
The answer to this equation is that the center of a circle (- a 2 + 0-2 + y) passes through the center of a circle (- a 2 + 0-1 + y), so it is symmetric
If point a is any point on the circle x2 + Y2 + ax + 4y-5 = 0 and the symmetric point of a on the line x + 2y-1 = 0 is also on the circle C, then the real number a is equal to______ .
Point a is any point on the circle x2 + Y2 + ax + 4y-5 = 0, and the symmetric point of a about the straight line x + 2y-1 = 0 is also on the circle C, which means that the straight line passes through the center of the circle, and the coordinates of the center of the circle (- A2, − 2) are substituted into the linear equation x + 2y-1 = 0, so − A2 − 4 − 1 = 0, so a = - 10, so the answer is: - 10
Point a is the point on the circle C: x ^ 2 + y ^ 2 + ax + 4y-5 = 0. If the symmetric point of a on the line x + 2Y = 0 is also on the circle C, then the real number a
Like the title,
in other words
(x + a) ^ 2 + (y + 2) ^ 2 = 1 + A ^ 2 the center of the circle is on X + 2Y = 0
The center coordinates are (a, 2), so a = 1
The solution set of known inequality ax ^ 6-3x + 2 > 0 is {x | XB}
1. Find the value of a and B; 2. Solve the inequality x ^ 2-B (a + C) x + 4C > 0
The title should be X & # 178;, the following is a change in the answer
1. Because the solution set of inequality is {x | XB}
So the two roots of ax & # 178; - 3x + 2 = 0 are 1 and B
Substituting 1 into a - 3 + 2 = 0, so a = 1
B & # 178; - 3B + 2 = 0, so the other root B = 2
2. Inequality X & # 178; - 2 (1 + C) x + 4C > 0
The opening is upward, so only △ 0 is needed
4（1 + c）² - 16c ＞ 0
c² - 2c + 1 ＞ 0
（c - 1）² ＞ 0
So the value range of C is C ≠ 1
It is known that the solution set of the inequality system x-a2 about X is 1 < x < 3. Find the solution set of the inequality ax-b > 0
x-a(b-a+2)/2
So (B-A + 2) / 20
X>1
From x-a (B-A + 2) / 2
And because the solution set is 1 < x < 3,
Therefore, we can get the equations: a + B + 1 = 3
(b-a+2)/2=1
The solution is a = b = 1
Therefore, we get the inequality: X-1 > 0
The solution is: x > 1
The solution set of ∵ x-a2 is 1 ∵ x ∵ 3
∴X＜a+b+1=3
X＞﹙2+b-a﹚/2=1
a=b=1
The solution set of ax-b ＞ 0
X＞1
Treat a and B as constant terms and solve the system of inequalities
The solution is: 1 + (B-A) / 2
The solution of inequality about X: ax & # 178; - (a + 2) x + 2 < 0 (a belongs to R)
ax²-（a+2）x+2＜0
(ax-2)(x-1)0,(x-2/a)(x-1)2,0
(x-1)(ax-2)1；
If a > 2, then 1 / A
If the solution set of inequality X & # 178; + AX-10 ＜ 0 is {x | - 5 ＜ x ＜ 2}, what is the value of a
That is, X & # 178; + AX-10 = 0 is followed by - 5 and 2
Then - 5 + 2 = - A
A=3
The inequality is regarded as an equation and multiplied according to the cross
We can get a = - 5 + 2 = - 3
（X+5）（X-2）＜0
A=3
It is known that the first term of arithmetic sequence (an) is a, the tolerance is B, and the solution set of inequality log2 (AX ^ 2-3x + 6) > 2 is (x ﹤ 1 or X ﹥ b)
1. Find the general term formula and the first n term and Sn formula of sequence (an)
2. Find the first n terms and TN of the sequence (1 / an multiplied by a n + 1)
1. The inequality log2 (AX ^ 2-3x + 6) > 2 is changed into ax ^ 2-3x + 6 > 4 and ax ^ 2-3x + 2 > 0, and its solution set is (x ﹤ 1 or x > b), ｜ 1 is the root of ax ^ 2-3x + 2 = 0, ｜ A-3 + 2 = 0, a = 1, ｜ B is the other root of x ^ 2-3x + 2 = 0, B = 2. ｜ an = 1 + 2 (n-1) = 2N-1.. Sn = n (1 + 2n-1) / 2 = n ^ 2.2.1 / [an *
In quadratic inequality of one variable, how to solve the quadratic coefficient as a parameter, such as: (a + 1) x & # 178; + 2x + 1 = 0
（a+1)x²+2x+1=0:
When a = - 1, the equation becomes 2x + 1 = 0, x = - 1 / 2
When a ≠ - 1, △ = 4-4 (a + 1) = - 4A,
When △ 0, the equation has no real root;
When △ = 0, i.e. a = 0, the equation has equal real root-1;
A > 0 is a