Let a, B, C be real numbers, f (x) = (x + a) (x2 + BX + C), G (x) = (AX + 1) (CX2 + BX + 1). Let s = {x | f (x) = 0, X ∈ r}, t = {x | g (x) = 0, X ∈ r}. If {s}, {t} are the number of elements of S, T & nbsp; respectively, then the following conclusion is impossible () A. {s} = 1 and {t} = 0b. {s} = 1 and {t} = 1C. {s} = 2 and {t} = 2D. {s} = 2 and {t} = 3

Let a, B, C be real numbers, f (x) = (x + a) (x2 + BX + C), G (x) = (AX + 1) (CX2 + BX + 1). Let s = {x | f (x) = 0, X ∈ r}, t = {x | g (x) = 0, X ∈ r}. If {s}, {t} are the number of elements of S, T & nbsp; respectively, then the following conclusion is impossible () A. {s} = 1 and {t} = 0b. {s} = 1 and {t} = 1C. {s} = 2 and {t} = 2D. {s} = 2 and {t} = 3

∵ f (x) = (x + a) (x2 + BX + C), when f (x) = 0, there is at least one root x = - A, when b2-4c = 0, f (x) = 0 has another root x = - B2, as long as B ≠ 2a, f (x) = 0 has two roots; when B = 2A, f (x) = 0 is a root; when b2-4c < 0, f (x) = 0 has only one root; when b2-4c > 0, f (x) = 0 has two or three roots. When a = b = C = 0, {s} = 1, {t} = 0, when a > 0, B = 0, C = 0 When ＞ 0, {s} = 1 and {t} = 1, when a = C = 1 and B = - 2, {s} = 2 and {t} = 2

Let a, B, C be a real number, f (x) (x + a) (x + 2 + BX + C), G (x) = (AX + 1) (AX ^ 2 + BX + 1), let the set s = {x \124124124124124124\, 124\124\124, \\\\124\\\\\\\124\124\\\\\\\124\\\\\\\| t | = 2D. | s | = 2 and | t | = 3, please analyze in detail

First of all, if we don't look at the option, we can know that f (x) = 0 and G (x) = 0 have at most three solutions, and f (x) = 0. We know that a solution is - A, but for G (x), such a solution may not exist, so for option a, if | t | = 0, then a = 0, otherwise g (x) = 0 has at least one solution, and once a = 0, but g (x) = 0, there is still no solution

Given the set a = {x | x2-3x + 2 = 0}, B = {x | x2 MX + M-1 = 0}, if a ∩ B = a, find the value range of real number M
It's the square of X

A = {x | x = 1 or x = 2},
A∩B=A :1-m+m-1=0 m∈R
2^2-2m+m-1=0 m=3
Therefore, the value range of real number m is: m ∈ R

Let a = {x | x2-3x + 2

We can get the solution set of a as 1 "X" 2,1. A is included in B, so we can find the value range of a (mainly because it is difficult to understand. We can draw a number axis.) so a is greater than or equal to 22. Anb = empty set, so we can find the value range of A. a must be less than or equal to 1. We sincerely hope that we can help the landlord