Let f (x) = (1 + a) x ^ 4 + x ^ 3 - (3a + 2) x ^ 2-4a（ Let f (x) = (1 + a) x ^ 4 + x ^ 3 - (3a + 2) x ^ 2-4a, try to prove that: for any real number a (1), the equation f (x) = 0 always has the same real root; (2) there exists X ', and f (x') ≠ 0

Let f (x) = (1 + a) x ^ 4 + x ^ 3 - (3a + 2) x ^ 2-4a（ Let f (x) = (1 + a) x ^ 4 + x ^ 3 - (3a + 2) x ^ 2-4a, try to prove that: for any real number a (1), the equation f (x) = 0 always has the same real root; (2) there exists X ', and f (x') ≠ 0

I'm glad to answer for you! F (x) = (1 + a) x ^ 4 + x ^ 3 - (3a + 2) x ^ 2-4a = (x ^ 4 + x ^ 3-2x ^ 2) + (AX ^ 4-3ax ^ 2-4a) = (x ^ 2 + X-2) x ^ 2 + a (x ^ 4-3x ^ 2-4) = (x + 2) (x-1) x ^ 2 + a (x ^ 2-4) (x ^ 2 + 1) = (x + 2) (x-1) x ^ 2 + a (x + 2) (x ^ 2 + 1) = (x + 2) [(x-1) x ^ 2 + a (X-2) (x ^ 2 + 1)]

Let a be a real number and a1 + I + 1 + I2 be a real number, then a = ()
A. 12B. 1C. 32D. 2

Let a be a real number, a1 + I + 1 + I2 = a (1 − I) 2 + 1 + I2 = (a + 1) + (1 − a) I2 be a real number, then a = 1, so choose B

It is known that the real number equation x2 + ax + B = 0 has two real numbers with α β, if | α|

It can be seen from the relationship between root and number
α+β=-a,αβ=b,
From | α|

Given that x > 1, we prove that x > LNX

Let f (x) = x-lnx, f (1) = 0, we need to prove that when x > 1, f (x) > F (1) = 0
Therefore, we only need to prove that f (x) increases when x > 1
F '(x) = 1-1 / x, when x > 1, obviously f' (x) > 0
Therefore, if f (x) increases on [1, positive infinity], then if x > 1, f (x) > F (1) = 0
That is x-lnx > 0, that is x > LNX