Equation 8 (x power of 4 + x power of 4) - 54 (x power of 2 + x power of 2) + 101 = 0

Equation 8 (x power of 4 + x power of 4) - 54 (x power of 2 + x power of 2) + 101 = 0

8 (4 ^ x + 1 / 4 ^ x) - 54 (2 ^ x + 1 / 2 ^ x) + 101 = 0 let 2 ^ x + 1 / 2 ^ x = t, then T ^ 2 = 4 ^ x + 1 / 4 ^ x + 24 ^ X-1 / 4 ^ x = T ^ 2-28 (T ^ 2-2) - 54t + 101 = 08T ^ 2-54t + 85 = 0 (4t-17) (2t-5) = 0t = 17 / 4 or T = 5 / 2 let 2 ^ x = u, then u + 1 / u = 17 / 4 or 5 / 2 if u + 1 / u = 17 / 4, then u = 4 or 1 / 4, then x = 2 or - 2 if u + 1 / u

Multiplication, division and factorization of integers
① 2A & # 178; + A ^ 8 △ a ^ 6 ② (- 2Ab) (- 1 / 3A & # 178; b) + 5ab × 1 / 3A & # 178; B ③ (X-Y + 9) (x + Y-9) ④ 2011 & # 178; - 2008 × 2014 ⑤ if (y + 3) (Y-2) = y & # 178; + my + N, get the value of Mn

①2a²+a^8÷a^6=2a²+a²=3a²②（-2ab）（-1/3a²b）+5ab×1/3a²b=2/3a³b²+5/3a³b²=7/3a³b²③（x-y+9）（x+y-9）=[x-(y-9)][x+(y-9)]=x²-(y-9)²...

Factorization A & #178; + 5ab + 4B

The last item of the title should be 4B;
a²+5ab+4b²
=(a+4b)(a+b)

(A's Square - 1) (A's square + 8A + 15) - 20, factoring

Original formula = (a + 1) (A-1) (a + 3) (a + 5) - 20
=[(a+1)(a+3)][(a-1)(a+5)]-20
=[(a²+4a)+3][(a²+4a)-5]-20
=(a²+4a)²-2(a²+4a)-15-20
=(a²+4a)²-2(a²+4a)-35
=(a²+4a+5)(a²+4a-7)

Factorization: [(a ^ 2) - A] ^ 2-8a ^ 2 + 8A + 12

The original form = [A & # 178; - A] &# 178; - 8 (A & # 178; - a) + 12
=[(a²－a)－2][(a²－a)－6)]
=(a²－a－2)(a²－a－6)
=(a－2)(a＋1)(a－3)(a＋2)

Given that function FX is an odd function on R, when x is greater than or equal to 0, FX = x (x + 1), if f (a) = - 2, find the value of real number a

If a > = 0
Then f (a) = a (a + 1) = - 2
a²+a+2=0
unsolvable
If A0
So f (- a) = (- a) (- A + 1)
For odd functions, then f (- a) = - f (a)
So f (a) = - f (- a) = a (- A + 1) = - 2
a²-a-2=0
a

Given that the domain of function FX is [- 2,3], find the domain of function f (x) + F (1-x)

A:
The domain of F (x) is [- 2,3]
Then the domain of F (x) + F (1-x) satisfies:
-2

Given f (x) = loga (a ^ x-1) (a > 0 and a ≠ 1), why is the value of x greater than 1

F (x) = log [a, a ^ X - 1] (a > 0 and a ≠ 1) find out why the value of X is greater than 1
When a > 1, a ^ X - 1 > a,
a^x > a + 1,
x > Log[a,a + 1],
When 0 < a < 1, 0 < A ^ X - 1 < A,
1 < a^x < a + 1,
Log[a,a + 1] < x < 0,

Why is FX even and GX odd, fx.gx It's an odd function
Is there a theorem
It's the multiplication of two functions

FX is an even function and GX is an odd function
∴f(-x)=f(x)
g(-x)=-g(x)
∴f(-x)g(-x)=f(x)×[-g(x)]=-f(x)g(x)
It is an odd function
It can be determined by definition

The function FX defined on R is odd, GX is even and FX GX = x-2x-3
Find the expression of FX and GX, so that FX = GX holds the value of X

Since f (x) is an odd function and G (x) is an even function, f (x) = - f (- x), G (x) = g (- x). According to the known conditions, we can get f (- x) - G (- x) = - x) ^ 2 + 2x-3, then - f (- x) - G (x) = x ^ 2 + 2x-3 and f (x) - G (x) = x ^ 2-2x-3 form a system of quadratic equations with two variables. We can get f (x) = - 2x, G (x) = - x ^ 2 + 3