The equation (x ^ 2-x + 1) / (x-1) = a + 1 / (a + 1) about X is transformed into the form of X + 1 / x = C + 1 / C, and the solution is ()

The equation (x ^ 2-x + 1) / (x-1) = a + 1 / (a + 1) about X is transformed into the form of X + 1 / x = C + 1 / C, and the solution is ()

(x^2-X+1)/(X-1)=[X(x-1)+1]/(X-1)=X+1/(X-1)
Minus one
（X-1）+1/(X-1)=（a+1）+1/(a+1)
Because the solution of X + 1 / x = C + 1 / C is x = C or x = 1 / C
So the solution of (x-1) + 1 / (x-1) = (a + 1) + 1 / (a + 1) is X-1 = a + 1 or X-1 = 1 / (a + 1)
X = a + 2 or x = 1 + 1 / (a + 1)
Blind change
The equation x & sup2; - x + 1 / X-1 = a + 1 / A-1 about X is transformed into the equation x = 1 / x = C + 1 / C
The equation x & sup2; - x + 1 / X-1 = a + 1 / A-1 about X is transformed into the equation x = 1 / x = C + 1 / C
The answer is X-1 + 1 / X-1 = A-1 + 1 / A-1
（x2-x+1）/x-1={x（x-1）+1}/（x-1）=X+1/（X-1）=a+1/a-1
If you subtract 1 from both sides, you get X-1 + 1 / (x-1) = A-1 + 1 / (A-1)
two hundred and twenty-two million five hundred and fifteen thousand five hundred and twenty-four
The equation x & sup2; - x + 1 / X-1 = a + 1 / A-1 about X is transformed into the equation x = 1 / x = C + 1 / C
The answer is X-1 + 1 / X-1 = A-1 + 1 / A-1
The transformation of equation x2-x + 1 / X-1 about x into equation a + 1 / A-1 is
(X²-X+1)/(X-1)=A+1/(A-1)
[x（x-1）+1]/（x-1）=A+1/(A-1)
X+1/(X-1)=A+1/(A-1)
X-1+1/(X-1)=A+1+1/(A-1).
If X & # 8308; - 5x & # 178; + 6 = 0, the original equation can be transformed into
x⁴-5x²+6=0
（x²-2）（x²-3）=0
Given that the tangent slope of the circle x ^ 2 + y ^ 2 = 9 is 3, find the tangent equation
Let y = 3x + B, that is, 3x-y + B = 0
Because it is tangent, the distance from the center of the circle to the straight line is equal to the radius
That is | 3 × 0-0 + B | / √ (3 ^ 2 + (- 1) ^ 2) = 3
Launch | B | = 3 √ 10
‖ B = 3 √ 10 or - 3 √ 10
The tangent equation is 3x-y + 3 √ 10 = 0 or 3x-y-3 √ 10 = 0
Suppose that the line is y = 3 * x + B
The combination of linear equation and circular equation can be solved, and after being substituted into the circular equation, we can get: 10x ^ 2 + 6bx + B ^ 2-9 = 0
Because it is tangent, the discriminant of root is 0, that is, 36B ^ 2-4 * (b ^ 2-9) = 0
So B ^ 2 = 9 / 8
B has two values, just root it. It's simple
It is proved by collocation method that no matter what real number x takes, the value of the algebraic formula - 2x & sup2; + 8x-18 is less than 0
Original formula = - 2 (x ^ 2-4x + 9) = - 2 (x ^ 2-4x + 4 + 5) = - 2 (X-2) ^ 2-10
Calculate (x ^ 2Y ^ 3) ^ 3 + (- 2x ^ 3Y ^ 2) ^ 2 * y ^ 5
(x^2y^3)^3+(-2X^3y^2)^2*y^5
=x^6y^9+4x^6y^4*y^5
=x^6y^9+4x^6y^9
=5x^6y^9
It is known that the univariate quadratic equation 2x square + 2mx + 2m-6 = 0 (m ∈ R) has real roots x ι, x2. (I) find the range of M. (II) Let f (m) = the square of X ι + the square of X ι * x2 + X2, find the maximum and minimum of F (m)
1. A = {- 1 / 2,1 / 3} x0d (2) f (m) has no maximum value, and the minimum value is 9. For the first question, find out X1 in P and bring x2 into Q. for the second question (1) Δ = 0, there is the value range of M. for the second question (2), find out the representation of M of X1 and x2. When f (m) is brought in, find out the minimum value when m = 1 / 2. Considering the value range of M, find out that when m = - 2, the minimum value is 9 and the maximum value is none
2Y (3-y) = 3 solved by the formula method of quadratic equation of two variables
2y(3-y)=3
6y-2y^2-3=0
2y^2-6y+3=0
The solution is as follows
Y = {6 ± radical [(- 6) ^ 2-4 * 2 * 3]} / (2 * 2)
=(6 ± root 12) / 4
=(3 ± root 3) / 2
Make the tangent of circle x ^ 2 + y ^ 2-4x = 0 through P (- 1, - 2), and find the tangent equation
There are three ways to solve this problem: 1. Let the linear equation of point p be y + 2 = K (x + 1), and then y + 2 = K (x + 1) and x ^ 2 + y ^ 2-4x = 0 together to form a system of equations. Eliminate y to get the quadratic equation of one variable about X, and then use discriminant = 0 to get how much K is equal to (this method is more troublesome, but it is the basic point of conic curve!) 2