Factorization in complex number: x square + 2x + 3 =? Don't just give me an answer, tell me how it came from

Factorization in complex number: x square + 2x + 3 =? Don't just give me an answer, tell me how it came from

x²+2x+3
=x²+2x+1+2
=(x+1)²+2
=(x+1+√2i)(x+1-√2i)

Factoring in complex numbers: 2x squared + X + 1. Please write the procedure

2x²+x+1=2[x²+(1/2)x]+1=2[x²+2×(1/4)x+(1/4)²]-1/8+1=2(x+1/4)²+7/8=2(4x+1)²/16+7/8=[(4x+1)²+7]/8=[(4x+1)²-(i√7)²]/8=(4x+1+i√7)(4x+1-i√7)/8

Factorization of x ^ 2-2x + 3 in complex numbers

x^2-2x+3
=(x-1-radical 2I) (x-1 + radical 2I)

The function f (x) = ACOS (2x π / 4) B, (x ∈ R) if the range of F (x) is [- 5,1]. Find the value of constant a, B and the monotone interval of function f (x)

If a > 0A + B = 1-A + B = - 5 = = > b = - 2; a = 3f (x) = 3cos (2x + π / 4) - 2, substitute 2x + π / 4 into the monotone increasing interval of standard cosine, and the solution is: - π + 2K π ≤ 2x + π / 4 ≤ 2K π = = > monotone increasing interval is: - 5 π

The opposite of the absolute value of 7 + A is - 2, then a=

That is to say, the absolute value of 7 + A is 2, then 7 + a = 2 or - 2, so a = - 5 or - 9

Range of logarithm function
The range of √ (4-x) - log2 (x)~
Don't answer if you don't know!

√ (4-x) is a decreasing function, - log2 (x) is also a decreasing function
The domain is (0,4)
So when the maximum value is x = 0, the minimum value is x = 4

Simplification of the absolute value of a blank filling question: first determine the number () in the absolute value sign, and then simplify according to the () of the absolute value
Such as the title~

Positive and negative, method

Let f (x) = LG (x + √ x ^ 2 + 1). (1) find the value of F (0) and (2) find the domain of F (x)

Substituting x = 0, we get f (0) = LG 1 = 0
X is under the root sign, so when x > = 0 and x > = 0, the true number is greater than 0, so the domain of definition is the left closed right open interval from 0 to positive infinity

It is known that the function f (x) = (AX + b) / (xsquare + 1) is an odd function defined on (- 1,1),
And f (1 / 2) = - 2 / 5, find the analytic expression of function f (x), judge the monotonicity, and solve the inequality f (t-1) + F (T) < 0

1、f(x)=(ax+b)/(1+x^2)
Because: F (x) is an odd function,
So: F (0) = b = 0, that is: F (x) = ax / (1 + x ^ 2)
And because f (1 / 2) = 2 / 5
So: a (1 / 2) / (1 + (1 / 2) ^ 2) = 2 / 5
That is: a (1 / 2) / (1 + 1 / 4) = a (2 / 5) = 2 / 5
So: a = 1
Therefore, the analytic formula is: F (x) = x / (1 + x ^ 2)
2. Let X1 < X2, and x1, X2 ∈ (- 1,1)
f(x2)-f(x1)=x2/(1+x2^2)-x1/(1+x1^2)
=[x2(1+x1^2)-x1(1+x2^2)]/[(1+x1^2)(1+x2^2)]
Obviously, if the denominator is greater than 0, we only need to examine the numerator
Molecule = x2 + X2 (x1 ^ 2) - x1-x1 (x2 ^ 2)
=(x2-x1)-x1x2(x2-x1)
=(x2-x1)(1-x1x2)
Because x1, X2 ∈ (- 1,1), so x1x2 < 1, that is: 1-x1x2 > 0
Because X1 < X2, x2-x1 > 0
So: when x2 > x1, f (x2) > F (x1)
That is: in the (- 1,1) domain, f (x) is an increasing function
3. Solving inequality f (t-1) + F (T) < 0
Because: F (x) = x / (1 + x ^ 2)
So the inequality becomes:
(t-1)/(1+(t-1)^2)+t/(1+t^2)＜0
[(t-1)(t^2+1)+t((t-1)^2+1)]/[(1+(t-1)^2)(1+t^2)]＜0
Because the denominator is greater than 0,
So (t-1) (T ^ 2 + 1) + T ((t-1) ^ 2 + 1) < 0
That is: 2T ^ 3-3t ^ 2 + 3t-1 < 0
t^3+(t-1)^3＜0
t^3-(1-t)^3＜0
Because T-1, t ∈ (- 1,1), so t ∈ (0,1)
So the above inequality becomes
t^3＜(1-t)^3
t＜1-t
2t＜1
t＜1/2
We have t ∈ (0,1),
Therefore, the solution of the inequality is as follows:
0＜t＜1/2

Given the function f (x) = a ^ (2x ^ 2-1), G (x) = a ^ (x ^ 2) (a > 0, and a ≠ 1), when x takes what value, f (x) > G (x)? Please have a clear process,

Because f (x) > G (x), then ln f (x) > ln g (x)
Then (2x ^ 2-1) ln a > x ^ 2 ln a
When a > 1
The solution of x ^ 2 - 1 > 0 is x > 1 or X