The solution of equation 4 to the power of X + 2 to the power of X-2 = 0 is

The solution of equation 4 to the power of X + 2 to the power of X-2 = 0 is

4^x+2^x-2=0
(2^x)^2+2^x-2=0
(2^x+2)(2^x-1)=0
2 ^ x + 2 = 0 or 2 ^ X-1 = 0
2 ^ x = - 2 (impossible)
2^x=1
x=0

The solution set of equation 4 with x power - 2 with x power = 0 is________

X power of 4-x power of 2 = 0
(2^x)²-2^x=0
2^x(2^x-1)=0
∴2^x-1=0
x=0

It is proved that the equation X5 + x3-1 has a unique real root between 0-1

When x = 0, the original formula = - 1 ＜ 0, when x = 1, the original formula = 1 ＞ 0, so at least one point of the original formula intersects with the X axis. The derivation of the original formula shows that 5x ^ 4 + 3x ^ 2 ≥ 0, so the original formula increases monotonically, so there is only one intersection point at most between the original formula and the X axis

Given that FX is an odd function, GX is an even function, and FX GX = 1-x2-x3, find FX GX

f(x)-g(x)=1-x^2-x^3
Substituting - x into the above formula, we get the following result:
F (- x) - G (- x) = 1-x ^ 2 + x ^ 3, that is - f (x) - G (x) = 1-x ^ 2 + x ^ 3
Add the two formulas and divide by 2 to get: - G (x) = 1-x ^ 2, get: G (x) = x ^ 2-1
The two formulas are subtracted and divided by 2 to get f (x) = - x ^ 3,

Given that FX is an odd function, GX is an even function, and FX + Gx = - x2 + 2x-3, then FX GX =?

f(x)+g(x)=-x^2+2x-3
-f(x)+g(x)=f(-x)+g(-x)=-x^2-2x-3
f(x)=2x
g(x)=-x^2-3
f(x)-g(x)=x^2+2x+3

Given the square of function FX = x + (a + 1) x + 2. A ≠ - 1. If FX = GX + HX, where GX is an odd function and HX is an even function, if the functions GX and FX are decreasing functions in the interval - infinity, 1], then the value range of real number a is

f(x)=g(x)+h(x)
f(-x)=g(-x)+h(-x)=-g(x)+h(x)
By subtracting the two formulas, G (x) = [f (x) - f (- x)] / 2 is obtained
So g (x) = (a + 1) X
G (x) in X

GX is a function defined on R, H (x) = f (x) + G (x), then "FX, GX are even functions" is "HX is even function"
What conditions, I know, are sufficient but not necessary? How to prove them?

Take a counter example
f(x)=x,g(x)=-x
Then: H (x) = 0
Obviously, H (x) is even, but f (x) and G (x) are odd
So, it's not necessary

If FX is an odd function, GX is an even function, FX GX = x square + 9x + 12, then FX + Gx

F (x) is an odd function, f (- x) = - f (x), G (x) is an even function, G (- x) = g (x)
f(x)-g(x)=x²+9x+12
f(-x)-g(-x)=(-x)²+9(-x)+12
-f(x)-g(x)=x²-9x+12
f(x)+g(x)=-x²+9x-12

It is known that F X is equal to the x power of ten and F X is equal to G x plus H x, where G x is an even function and H x is an odd function to find G x and H X

g(x)=g(-x)  h(x)=-h(-x)
f(x)=g(x)+h(x)=10^x
      f(-x)=g(-x)+h(-x)
            =g(x)-h(x)=10^-x
g(x)=2/(10^x+10^-x)
h(x)=2/(10^x-10^-x)

On the domain of odd and even functions
Must the domain of definition of odd function be symmetric about the origin?
Even function? Is it the same?

Yes