1. (1 in 100 x 1 in 99 x 1 in 98 X 3 / 1 x 2 / 1 x 1) to the power of 200 (100x99x98x97 X3x2x1) to the power of 200 2. (2nth power of B) to the third power (the third power of B) to the fourth power (the fifth power of B) n + 1 power 3. {0.4x-0.2y = 3.9 and 3 / 5-5 / 5 4Y = 1.7}

1. (1 in 100 x 1 in 99 x 1 in 98 X 3 / 1 x 2 / 1 x 1) to the power of 200 (100x99x98x97 X3x2x1) to the power of 200 2. (2nth power of B) to the third power (the third power of B) to the fourth power (the fifth power of B) n + 1 power 3. {0.4x-0.2y = 3.9 and 3 / 5-5 / 5 4Y = 1.7}

Using (x1 * x2 *... Xn) ^ m = X1 ^ m * x2 ^ m *... Xn ^ m
1 / 100 is paired with 100, and so on, the result is 1
Use (x ^ m) ^ n = x ^ (m * n)
The original formula = B ^ (6N + 12n-5 * (n + 1)) = B ^ (13n-5)
The equation can be solved by elimination method
The original formula - & gt; 4x-2y = 39 (1); 3x-4y = 8.5 (2) (reduction coefficient) - & gt; 2 * (1) - (2) 5x = 69.5, x = 13.9 - & gt; y = 8.3

Which is the greater of the sixth power of X + 1 or the fourth power of X + the second power of X

Difference method
x^6-x^4-x^2+1
=(x^6-x^4)-(x^2-1)
=x^4(x^2-1)-(x^2-1)
=(x^4-1)*(x^2-1)
=(x^2+1)(x^2-1)^2>=0
So when x = + - 1, the sixth power of X + 1 is equal to the fourth power of X + the second power of X
In other cases, the sixth power of X + 1 is greater than the fourth power of X + the second power of X

4 times 2 to the power of x times 6 = 2 to the power of 13

A:
4 times 2 to the power of x times 6 = 2 to the power of 13
4×(2^x)×(6^x)=2^13
(2^2)×(2×6)^x=2^13
12^x=2^(13-2)=2^11
Logarithm of two sides:
xln12=11ln2
So: x = 11ln2 / ln12

X-6 X's bisection + 1 times X's third power + X's bisection-36
It's urgent. Five minutes

=(x²+1)/(x-6)×(x+6)(x-6)/x(x²+1)
=(x+6)/x