If (2x-3) = (2x-1 / a) plus (2x + 1 / b), then the values of a and B are

If (2x-3) = (2x-1 / a) plus (2x + 1 / b), then the values of a and B are

Combine the scores on the right side of the equation with the scores to get ((2a + 2b) x + a-b) / (4x ^ 2-1)
Because it is equal to the left side of the equation, we get
2A+2B=2
A-B=-3
The solution is a = - 1
B=2
A equals - 1, B equals 2
(2x-3)/(4x^2-1) = A/(2x-1)+B/(2x+1) ==> (2x-3)/((2x-1)*(2x+1)) = A/(2x-1) + B/(2x+1)
==> 2x-3 = A*(2x+1) + B*(2x-1) ==> 2x-3 = 2Ax + A + 2Bx - B = (A+B)*2x + (A-B)
==>A + B = 1 and A-B = - 3 = = > A = - 1 and B = 2
First merge, get ((2a + 2b) x + a-b) / (4x ^ 2-1)
Because it is equal to the left side of the equation, we get
2A+2B=2
A-B=-3
The solution is a = - 1
B=2
When m takes what value, 2x-3 / X-2 = m + 4 / 2-x + 1 will produce increasing roots
Is your topic right
（2x-3）/（x-2）=（m+4）/（2-x）+1
-2x+3=m+4+2-x
m=-x-3
X = 2 is the increasing root of the equation
At this point, M = - 5
So when m = - 5, there will be increasing roots
When m takes what value, the equation 2x-1 parts x + 3 = (2x-1) (7-2m) parts (x + 3) (3m + 2) holds?
emergency
When the equation is (2x + 2 m) (x = 1-3 m)
∴7-2m=3m+2;
5m=5;
m=1;
I'm very glad to answer your questions. Skyhunter 002 will answer your questions
If you don't understand this question, you can ask,
If M > 25, find the solution set of inequality 5mx < 2x - 1
The inequality can be reduced to X (5m-2) ＜ - 1, ∵ m ＞ 25, ∵ 5m-2 ＞ 0, then x ＜ - 15m − 2
It is proved that the value of the algebraic formula 2x-x + 3 is not less than 23 / 8
2X ^ 2-x + 3 = 2 (x-1 / 4) ^ 2-1 / 8 + 3 = 2 (x-1 / 4) ^ 2 + 23 / 8 ≥ 23 / 8 when x = 1 / 4, get the equal sign
The equation of a line with a distance of 3 from the line 3x-4y + 1 = 0 is ()
A 3x-4y + 4 = 0 or 3x-4y-12 = 0
B 3x-4y + 16 = 0 or 3x-4y-14 = 0
How is that?
B is calculated as follows:
Two equal lines
The distance formula from 3x-4y + 16 = 0 to 3x-4y + 1 = 0 is as follows:
16-1 / radical (3 ^ 2 + (- 4) ^ 2) = 3
The teacher should say this formula in class
Given that x1, X2 are two real roots of the equation (A-1) x2 + X + A2-1 = 0, and X1 + x2 = 13, then x1 · x2=______ .
∵ X1 + x2 = 13, ∵ 1a − 1 = 13, the solution is a = - 2, then A2 − 1a − 1 = 4 − 1 − 2 − 1 = - 1, ∵ x1 · x2 = - 1
Given x-2y = 2, find the value of X & # 178; - 4xy + 4Y & # 178; - x + 2Y
x-2y=2
The original formula is (y-2x) - 2x
=2^2-2
=2
We know that the equation of circle is x ^ 2 + y ^ 2 + ax + 2Y + A ^ 2 = 0, and a certain point is a (1,2). If we want to make two tangents of circle through fixed point a, we can find the value range of A
Detailed process and thinking process, thank you
Because if you want to have two tangents, point a must be outside the circle, and one point must be outside the circle. The equation brought into the circle is greater than zero
1 ^ 2 + 2 ^ 2 + A + 4 + A ^ 2 ＞ 0 solve the inequality: A is all real numbers
It's a circle ^ 2 + 2, because it's a circle,
Therefore, d ^ 2 + e ^ 2-4f > 0
So: A ^ 2 + 4-4a ^ 2 > 0
So a ^ 2
There are two tangents to make a circle through point a,
So a is outside the circle
x^2+y^2+ax+2y+a^2=0
(x+a/2)^2+(y+1)^2=1-3a^2/4
Center O (- A / 2, - 1)
Then Ao > radius
That is Ao ^ 2 > R ^ 2
(1+a/2)^2+(2+1)^2>1-3a^2/4
1+a+a^2/4+9>1-3a^2/4
a^2+a+9>0
The inequality is expanded
There are two tangents to make a circle through point a,
So it's on the outside of A
x^2+y^2+ax+2y+a^2=0
(x+a/2)^2+(y+1)^2=1-3a^2/4
Center O (- A / 2, - 1)
Then Ao > radius
That is Ao ^ 2 > R ^ 2
(1+a/2)^2+(2+1)^2>1-3a^2/4
1+a+a^2/4+9>1-3a^2/4
a^2+a+9>0
This inequality is always true
So as long as R ^ 2 = 1-3a ^ 2 / 4 > 0
3a^2/4
When x=__ The maximum of the algebraic formula - 2x & sup2; + 3x-3___ The value is___ It is proved that the value of the algebraic formula 3x & sup2; - 6x + 5 is not less than 2
It is proved by the collocation method that the equation (3m & sup2; - 12m + 22) x & sup2; + 3mx + 1 = 0 is a quadratic equation with one variable no matter what the value of M is
When x = 3 / 4, the maximum value of - 2x & sup2; + 3x-3 is - 15 / 8
3x²-6x+5 = 3(x²-2x+1-1)+5 = 3[(x-1)²-1]+5 = 3(x-1)²+2
∵ 3 (x-1) & sup2; ≥ 0 ∵ 3 (x-1) & sup2; + 2 ≥ 2, that is, the value of the algebraic expression 3x & sup2; - 6x + 5 is not less than 2
It is proved by the collocation method that the equation (3m & sup2; - 12m + 22) x & sup2; + 3mx + 1 = 0 is a quadratic equation with one variable no matter what the value of M is
The quadratic coefficients of the equation 3M & sup2; - 12m + 22 = 3 (m-2) & sup2; + 10 ≥ 10
Whatever the value of M, the coefficient of quadratic term is not zero
The equation is a quadratic equation of one variable