Given that a > 0, b > 0, prove that a ^ n + B ^ n ≥ a ^ (n-1) B + AB ^ (n-1) n > 1, n belongs to Z

a^n+b^n-（a^(n-1)b+ab^(n-1) ）
=(a-b) (a ^ (n-1) - B ^ (n-1)) will be discussed in three cases
1, a > b > 0, then from the inequality product rule, a ^ (n-1) > b ^ (m-1) > 0, A-B > 0, so there is (a-b) (a ^ (n-1) - B ^ (n-1)) > 0
2, a = B, then (a-b) (a ^ (n-1) - B ^ (n-1)) = 0;
3,0

Prove that: (2 + radical 3) ^ n can always be expressed as a + B radical 3, where N, a and B are positive integers
Prove by mathematical induction (don't omit the key steps)

When n = 1, do it yourself
If n = k, then: (2 + radical 3) ^ k can be expressed as a + B radical 3, a and B are positive integers
Then when n = K + 1,
(2 + radical 3) ^ (K + 1) = (a + B radical 3) * (2 + radical 3) = 2A + a radical 3 + 2B radical 3 + 3B
=(2a + 3b) + (a + 2b) radical 3
Because a and B are positive integers, 2A + 3b and a + 2B are both positive integers
So

Even function f (x) defined on [- 3,3], when x ≥ 0, f (x) is an increasing function, if f (1-m) > F (m) holds, the value range of M is obtained
Well, copy it in the wrong book,

Even function, the axis of symmetry is x = 0, and it is an increasing function on the positive half axis, so the point closer to x = 0 is smaller
So there are: | 1-m | > | m | - > (1-m) ^ 2 > m ^ 2 - > M

I want to buy some clothes?
2. Is the noun clothes singular or plural? I know clothe means cloth
Or is it the general term of clothes, which belongs to the plural

1 I want to buy some clothes
The noun "clothes" is a collective noun, just like family. The predicate verb should be plural

How many meters is four foot two

1.4m

A number whose square equals nine fourths is zero

Plus or minus three thirds

English translation

who is your favourite sports man
Walking is my grandfather's favourite sport.
I can ride a bike but can not drive a car

It is proved that the product of any three consecutive positive integers must be divisible by 3
No mathematical induction

Let the three numbers be (n-1), N, (n + 1) (n is a positive integer greater than 2), then the product s = (n + 1) (n-1) n = (n * n-1) n = n * n * n-n. if n is divided by 3, then the remainder of s divided by 3 is 1 * 1 * 1-1 = 0. If n is divided by 3, then the remainder of s divided by 3 is 2 * 2 * 2-2 = 6. If n is a multiple of 3, then it is obviously divisible by 3

Let z = Z (x, y) be determined by the equation x ^ 2 + Z ^ 2 = y * f (Z / y), and find the partial Z / partial x (where f is a differentiable function)

The partial derivation of X from equation 2 is obtained
2 * x + 2 * Z * (partial Z / partial x) = y * (partial f / partial x) * (partial Z / partial x) / Y
So: partial Z / partial x = (2 * x) / [(partial f / partial x) - 2 * Z]
Note: in the result of partial f / partial x, the expression is still a function of (Z / y)

Please answer in detail, thank you! (18 16:52:49)
"All kinds of & # 160; - # 160; - # 160; - # 160; - # 160; - # 160; - # 160; - # 160; - # 160; - # 160; - # 160; in English?
Urgent~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

All kinds of
A wide range of

Given that a > 0, b > 0, prove that a ^ n + B ^ n ≥ a ^ (n-1) B + AB ^ (n-1) n > 1, n belongs to Z