F (x) = ax + 1 / ax + B How to find the derivative of this

F (x) = ax + 1 / ax + B How to find the derivative of this

f'(x)=a-1/ax^2
If the constant is 0,1 / x, it can be regarded as x ^ (- 1), which still satisfies the rule of polynomial derivation~

1 / 2 minus 5 / 6 minus 11 / 12 minus 19 / 20 minus 29 / 30
Know the reward of wealth 200

=1-2:1 + 1-6:1 + 1-12:1 + 1-20:1 + 1-30:1
=1 * 5 - (1-2 / 1 + 2 / 1-3 / 1 + 3 / 1-4 / 1 + 4 / 1-5 / 1 + 6 / 1)
=5 - (1-1 / 6)
=5-5 / 6
=4 and 1 / 6

The product of a fraction multiplied by 4 / 5, 2 / 3, 1 / 4 is a positive integer. What is the minimum of this fraction?

4 / 5 * 2 / 3 * 5 / 4 = 2 / 3 2 / 3 * 3 / 2 = 1, so = 3 / 2

We know that K is a positive integer, and prove that: (1) if K is the product of two continuous positive integers, then 25K + 6 is also two continuous positive integers

Let k = m (M + 1)
25K + 6 = 25m (M + 1) + 6 = (5m + 2) (5m + 3) is the product of two continuous positive integers 5m + 2 and 5m + 3

Let a positive integer be the product of two three digit positive integers. If the two three digit numbers are written together as a six digit number, then it is 7 times of the original product. Then the two three digit numbers are____ .

Let the two three digit numbers be x and y
1000X+Y=7XY
Y=1000X/（7X-1）=143+（143-X）/（7X-1）
Obviously 143-x

Known: the sum and product of three positive integers are equal, find these three positive integers!

a+b+c=abc
Let a ≥ B ≥ C
If C ≥ 2, then a ≥ 2, B ≥ 2, so AB > A + B, BC > b + C, AC > A + C = > ABC > (a + b) C > A + B + C
So C = 1
So AB = a + B + 1 = > (A-1) (B-1) = 2
Because a > b is a positive integer
So A-1 = 2, B-1 = 1 = > A = 3, B = 2, C = 1

Calculation: 1) square of 11 + square of 12 +... + square of 19 2) square of 1 + 1 / 4 + 1 / 4 +... + 1
calculation:
1) Square of 11 + square of 12 +... + square of 19
2) Square of 1 + 1 / 4 + 1 / 4 +... + 100th power of 1 / 4

1、1+2^2+3^2+...+n^2=n(n+1)(2n+1)/6,
——》Original formula = (1 + 2 ^ 2 + 3 ^ 2 +... + 19 ^ 2) - (1 + 2 ^ 2 + 3 ^ 2 +... + 10 ^ 2)
=(19*20*39-10*11*21)/6=2085；
2、1+q+q^2+...+q^n=1*[1-q^(n+1)]/(1-q),
——》The original formula = [1 - (1 / 4) ^ 101] / (1-1 / 4) = 4 / 3-1 / 3 * 4 ^ 100

11 to 19 square

11 square = 121 18 square = 324
12 square = 144 19 square = 361
Square of 13 = 169
14 squared = 196
Square of 15 = 225
Square of 16 = 256
Square of 17 = 289

12 (x + y) square + 11 (X-Y) + 2 (X-Y) square
Factorization

12 (x + y) square + 11 (X-Y) + 2 (X-Y) square
=12x²+24xy+12y²+11x²-11y²+2x²-4xy+2y²
=25x²+20xy+3y²
=（5x+y）（5x+3y）
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If the square of | X-2 | + (y + 0.25) is 0, find the x power of Y

If one is greater than 0, the other is less than 0
So both are equal to zero
So X-2 = 0, y + 0.25 = 0
x=2,y=-0.25
So the x power of y = (- 0.25) & sup2; = 0.0625