Solving the fractional equation: X-1 / x + 2 = x-3 / x + 4

Solving the fractional equation: X-1 / x + 2 = x-3 / x + 4

X-1/X+2=X-3/X+4
The result of cross multiplication is as follows
(x+2)(x-3)=(x-1)(x+4)
x^2-x-6=x^2+3x-4
4x=-2
x=-1/2.
After testing, x = - 1 / 2 is the root of the equation

[ellipse] known equation (3m + 7) x ^ 2 + (3m + 4) y ^ 2 = 5m + 12
The curve is ellipse, and the value range of real number m is calculated

According to the meaning: A ^ 2 = (5m + 12) / (3m + 7), B ^ 2 = (5m + 12) / (3m + 4)
c^2=a^2-b^2=(5m+12)/(3m+7)-(5m+12)/(3m+4)=(-15m-36)/(3m+7)(3m+4)
∵ the curve is ellipse, ∵ the eccentricity is less than 1
∴e=c/a=√[(-15m-36)/(3m+7)(3m+4)]/√[(5m+12)/(3m+7)]＜1
You can calculate the specific answer yourself

Let the equation of circle C1 be (x + 2) ^ 2 + (y-3m-2) ^ 2 = 4m ^ 2, and the equation of line l be y = x + m + 2
(1) Find the equation of C1 with respect to l-symmetric circle C2;
(2) When m changes and m ≠ 0, we prove that the center of C 2 is on a fixed line, and find the common tangent equations of a series of circles represented by C 2
Thank you for your reply on the first floor!

Center O (- 2,3m + 2)
Let the symmetric point of O with respect to l be B (a, b)
Then the line ob is perpendicular to L and the midpoint of ob is on L
The slope of L = 1
So ob slope (3m + 2-B) / (- 2-A) = - 1
3m+2-b=a+2
a+b=3m
The midpoint of OB [(A-2) / 2, (3m + 2 + b) / 2] is on L
Then (3m + 2 + b) / 2 = (A-2) / 2 + m + 1
a-b=m+2
a+b=3m
So a = 2m + 1
b=m-1
The radii of two circles are equal
So it's (x-2m-1) ^ 2 + (y-m + 1) ^ 2 = 4m ^ 2