If the solution set of inequality (M + 1) x > m + 1 about X is X

If the solution set of inequality (M + 1) x > m + 1 about X is X

Divide both sides by M + 1
The direction of unequal sign of solution set
So m + 1 is negative
So + 1

Inequality ax ^ 2 + ax + (A-1)

When a = 0 and there is - 10, the opening of the parabola ax ^ 2 + ax + (A-1) is upward. Obviously, the solution set of the inequality cannot take all the real number sets
When a

If the inequality (m-1) x ^ 2 + 2mx + m > 0 holds for any real number x, the value range of real number M

1.m=1
2x+1>0
Not always established
2.m≠1
M-1 > 0, that is, M > 1
△=4m²-4m（m-1）

The circle equation is (x-1) ^ 2 + (Y-1) ^ 2 = 1, and the point P coordinate is (2,3). The tangent equation of the circle passing through point P is obtained

Center (1,1), radius r = 1, distance from center to tangent is equal to radius. If tangent slope does not exist, then it is perpendicular to X axis. If passing through P, then x = 2 (1,1) to x = 2 distance = | 1-2 | = 1 = R, so x = 2 is tangent. If tangent slope exists, then Y-3 = K (X-2) kx-y-2k + 3 = 0 (1,1) to tangent distance = | k * 1-1-2k + 3 | / √ (k ^ 2 + 1) = 1 | K-2 | =

The equation of a circle in the mathematics of grade one of senior high school. Given the equation of a circle and a point outside the circle, how can we find the tangent equation of the circle at this point?
It's better to have an example to explain. It's better to have all kinds of methods

If you haven't learned derivative, follow these steps:
1, set the tangent point as (x0, Y0)
2. Substituting the tangent coordinates into the equation of the circle
3, denote the slope of the radius passing through the tangent point and the slope of the tangent line, so that the product of the two is - 1
4. Solve the equations obtained from 2 and 3 to get the tangent point coordinates
5. Get the tangent equation from the two-point formula

Two tangent lines PA, Pb of circle x2 + y2 = 10 from moving point P
The slopes of the two tangent lines PA, Pb and the straight line PA and Pb leading from the moving point P to the circle x2 + y2 = 10 are K1 and K2 respectively
(1) If K1 + K2 + K1 × K2 = - 1, find the trajectory equation of the moving point P
(2) If the point P is on the straight line x + y = m and AP ⊥ BP, find the value range of real number M

1) If K1 + K2 + K1 × K2 = - 1, find the trajectory equation of the moving point P
Let p be (a, b),
The straight line is y-b = K (x-a)
Substituting into the circular equation
x²+(kx-ak+b)²=10
(1+k²)x²-2kx(ak-b)+(ak-b)²-10=0
Because a line is tangent to a circle, the equation has only one real root
Then 4K & sup2; (ak-b) & sup2; = 4 (1 + K & sup2;) [(ak-b) & sup2; - 10]
a²k^4-2abk³+b²k²=a²k^4-2k³ab+k²(b²-10)+a²k²-2abk+b²-10
(a²-10)k²-2abk+b²-10=0
Then K1 + K2 = 2Ab / (A & sup2; - 10), K1 * K2 = (B & sup2; - 10) / (A & sup2; - 10)
Because K1 + K2 + K1 × K2 = - 1,
Then 2Ab / (A & sup2; - 10) + (B & sup2; - 10) / (A & sup2; - 10) = - 1
2ab+a²-10+b²-10=0
(a+b)²=20
The trajectory of point P is x + y = ± 2 √ 5, except for two points (± √ 5, ± √ 5)
2) If point P is on the straight line x + y = m and AP ⊥ BP, find the value range of real number M
It has been proved that K1 * K2 = (B & sup2; - 10) / (A & sup2; - 10)
AP⊥BP
Then K1 * K2 = - 1
Then (B & sup2; - 10) / (A & sup2; - 10) = - 1
a²+b²=20
The trajectory of point P is X & sup2; + Y & sup2; = 20
There is P on the line x + y = M
Then (M-Y) & sup2; + Y & sup2; - 20 = 0
y²-my+m²/2-10=0
Then M & sup2; > = 4 (M & sup2 / 2-10)
m²

When a_____ The square of fraction 2a-4-a is - 5A + 6
Square of x-4y square of X + 4xy + 4Y divided by square of X + xy-6y square of X + 5xy + 6y square of X

If you haven't finished the first question? (2a-4) / (a ^ 2-5A + 6) = 2 (A-2) / [(A-2) (A-3)] = 2 / (A-3) a = 2 or a = 3, the fraction is meaningless; if a ≠ 2 and a ≠ 3, the fraction is meaningful, depending on whether your question asks these questions

Solution set of equation (x ^ 2-1) (x ^ 2 + 2x-8) = 0

{1,-1,2,-4}

Solution set of equation (X-2) (x ^ 2-2x + 1) = 0

(x-2)(x^2-2x+1)=0
It can be reduced to (X-2) (x-1) ^ 2 = 0
So X-2 = 0 or X-1 = 0
therefore
x1=2,x2=x3=1

The set of solutions of equation X-1 = 0

{-1,1}