When m is a value, the equation X-2 / (x-3) = MX / (x-3) + 2 has no solution

When m is a value, the equation X-2 / (x-3) = MX / (x-3) + 2 has no solution

If x (x-3) - 2 = MX + 2 (x-3) is reduced to x ^ 2 - (M + 5) x + 4 = 0, then Δ

If the equation LX + 3-1 = mxx + 3 about X has no solution, then the value of M is ()
A. - 13b. - 1C. - 13 or - 1D. Cannot be determined

If M + 1 = 0, i.e. M = - 1, the equation has no solution; if M + 1 ≠ 0, i.e. m ≠ - 1, according to the meaning of the question, we get: x + 3 = 0, i.e. x = - 3, substituting x = - 3 into the integral equation, we get: - 3M = 1, the solution is: M = - 13, to sum up, the value of M is - 13 or - 1

It is known that the equation X4 + 2x3 + (3 + k) x2 + (2 + k) x + 2K = 0 has real roots, and the product of all real roots is - 2, then the sum of squares of all real roots is______ ．

This is the case of (x4 + 2x3 + 2x3 + 2x3 + (3 + k) x (2 + k) x (2 + k) x (2 + k) x (2 + k) x (2 + k) x2 + (2 + k) x (2 + k) x (2 + k) x (2 + 2x3 + 2x3 + (3 + k) x (2 + k) x (2 + k) x (2 + k) x (2 + k) x + 2 + 2 (2 + 2 + k) (2 + 2 + k) (2 + 2 + k) (2 + 2 + 2 + k) (2 + 2 + 2 + X (2 + 2 + k) (2 + 2 + 2 + 2 + k) (2 + 2 + 2 + X (2 + 2 + k) (2 + 2 + 2 + 2 + k) (2 + 2 + 2 + X (2 + 2 + 2 + k) (2 + 2 + X (2 + 2 + 2 + k) (2 + 2 + 2 + X (2 + 2 + 2 + 2 + x) (x) (x + k = 0, ∵ equation X4 + 2x3 + (3 + k) x2 + (2 + k) x +2K = 0, the product of all real roots is - 2, that is, the solution of the real roots of the original equation is equivalent to x2 + X-2 = 0, the two real roots are - 2 and 1, and the sum of squares of all real roots = (- 2) 2 + 12 = 5

Given that the equation 4K (x + 2) - 1 = 2x about X has no solution, find the value of K

4k（x+2）-1=2x
4KX+8K-1-2X＝0
（4K-2）X+8X-1＝0
（4K-2）X＝1-8K
X＝（1-8K)/（4K-2）
When 4k-2 = 0, the equation has no solution
When k = 1 / 2, the equation has no solution