7x ^ 4 * x ^ 5 * (- x) ^ 7 + 5 (x ^ 4) ^ 4 - (x ^ 8) ^ 2

7x ^ 4 * x ^ 5 * (- x) ^ 7 + 5 (x ^ 4) ^ 4 - (x ^ 8) ^ 2

7x^4*x^5*(-x)^7+5(x^4)^4-(x^8)^2
=-7x^9*x^7+5x^8-x^16
=-7x^16+5x^8-x^16
=-8x^16+5x^8

10x-3 =7x+3

10x-3 =7x+3
10x-7x=3+3
3x=6
x=2

7x = 10x + 6 to solve the equation

3X=-6 x=-2

Can you help me solve the problem of exponential function
Given that a > 0, the x power of F (x) = 3 divided by a + the x power of a divided by 3 is an even function on R, find the value of A

f(x)=3^x/a+a/3^x
f(-x)=1/3^xa+a3^x
f(x)=f(-x)
Then a = 1 / A and a > 0
So a = 1

A (X & # 178; + x) + B (X & # 178; - x) = 1-C write out the general form, find out the number of quadratic terms, the number of primary terms and constants, and prove that it is a quadratic equation of one variable

(a+b)x^2+(a-b)x+c-1=0
The quadratic term is (a + b) x ^ 2
The first term is (a-b) X
Constant C-1

1) If the function f (x) = a ^ x (a ^ X - 3A ^ 2 - 1) (a > 0 and a ≠ 1) is an increasing function in the interval [0, + ∞), then the value range of real number a is?
2) Let a, B, C be positive numbers, and 3 ^ a = 4 ^ B = 6 ^ C, then 2 / C = 2 / A + 1 / B, please prove!
3) Given 2 ^ A * 5 ^ B = 2 ^ c * 5 ^ D = 10, prove: (A-1) (D-1) = (B-1) (C-1)
Explanation: I want the specific process, and refuse to talk nonsense!

1. Let a ^ x = t, then
y=t(t-3a²-1)=[t²-(3a²+1)t+(3a²+1)²/4]-(3a²+1)²/4
=[t-(3a²+1)/2]²-(3a²+1)²/4
When t ≥ (3a & sup2; + 1) / 2, f (x) increases monotonically
That is, a ^ x ≥ (3a & sup2; + 1) / 2
That is, xlna ≥ ln (3a & sup2; + 1) - LN2
That is, X ≥ [ln (3a & sup2; + 1) - LN2] / LNA
∵x∈[0,+∞)
∴[ln(3a²+1)-ln2]/lna≤0
When a > 1, LNA > 0
That is, 3A & sup2; + 1 ≤ 2, the solution is - √ 3 / 3 ≤ a ≤ √ 3 / 3 (rounding off)
When 0

The quadratic equation of one variable (x-3) ² = 4 is reduced to the general form of (x-3) ² = 4____ The quadratic term is____ , the number of items at a time is____ -The constant term is
It is known that the quadratic equation (m-1) x & # 178; + X + 1 = 0 with respect to X has real roots, then the value range of M is sharp

Solution
(x-3)²=4
x²-6x+9=4
X & # 178; - 6x + 5 = 0 -- general formula
The quadratic term is: X & # 178;
The first term is: - 6x
The constant term is: 5
The equation has real roots
∴△=b²-4ac=1-4(m-1)≥0
And M-1 ≠ 0
That is, 1 ≥ 4 (m-1) and m ≠ 1
∴m-1≤1/4
Ψ m ≤ 5 / 4 and m ≠ 1

Senior one on the definition of exponential function and range of problems!
What is the domain of definition and the range of value under the root of y = 5 to the power of 3x-2?
The exponent of 5 is 3x-2 under the root

Is it the 3x-2 power of y = (root sign 5)? In this case... A ∈ (0, + ∞), D ∈ R or the - 2 power of y = 5 (root sign 3x)? In this case... A ∈ (- ∞, 0) ∪ (0, + ∞), D ∈ (- ∞, 0) ∪ (0, + ∞)... Or I don't understand it right. I hope LZ can express it clearly

The relationship between the root and coefficient of the first order equation of two variables
First read the solution of the first question, and then explore the second question
1 given that P ^ 2-p-3 = 0, 1 / Q ^ 2-1 / q-3 = 0, P, q are real numbers, and PQ ≠ 1, find the value of P + 1 / Q
∵pq≠1∴p≠1/q
And ∵ p ^ 2-p-3 = 0,1 / Q ^ 2-1 / q-3 = 0
P, 1 / Q are two unequal real roots of quadratic equation with one variable
From the relationship between root and coefficient, P + 1 / Q = - (- 1) = 1
2 given 2m ^ 2-3m-7 = 0, 7n ^ 2 + 3n-2 = 0, m, n are real numbers, and Mn ≠ 1, find the value of M = 1 / n
(hope there is a process)

∵mn≠1∴m≠1/n
And ∵ 2m ^ 2-3m-7 = 0,
7n^2+3n-2=0 7+3/n-2/n^2=0 2*(1/n)^2-3*(1/n)-7=0
M, 1 / N are two unequal real roots of quadratic equation with one variable
From the relationship between root and coefficient, m * (1 / N) = - 7 / 2

Give the following four exponential functions: y = (1 / 2) ^ x, y = 2 ^ x, y = (1 / 3) ^ x, y = 3 ^ x, if (1 / 2) ^ X1 = 2 ^ x2 = (1 / 3) ^ X3 = 3 ^ X4 = 3 / 2, try to arrange x1, X2, X3, X4 from large to small

1. According to (1 / 2) ^ X1 = (1 / 3) ^ X3: x1x4
3. According to (1 / 3) ^ X3 = 3 ^ X4: X4 > 0.x3x3
To sum up: x2 > X4 > X3 > x1