If 0 is greater than a and greater than 1, then a, one of a, the square of a, how to arrange from small to large? Sorry, it should be "0 less than a, less than 1". Sorry

If 0 is greater than a and greater than 1, then a, one of a, the square of a, how to arrange from small to large? Sorry, it should be "0 less than a, less than 1". Sorry

The square of a, a, one of A

a. - A, the square of a, - A, one part of a, one part of a (a is greater than zero), please arrange from small to large

When a ＞ 1, a & # 178; ＞ a ＞ one part of a ＞ A & # 178; one part ＞ - a ＞ - A & # 178;
When a is equal to 1, all are equal
When 0 ＜ a ＜ 1, a & # 178; one part ＞ one part ＞ a ＞ A & # 178; > - A & # 178; > - A

A = - 0.3 square, B = - 3-2 power, C = 1, d = 3-2 power, please order from small to large

a=0.09 b=-1/9 c=1 d=1/9
∴b

0 "X" 1, then the square of X, x, 1 / 2 of X are arranged from small to large as:
=With a few

X & sup2; < x < x Branch 1

If (- AB) × (- AB) × (- AB) > 0, then () a.ab0 C.A > 0,
If (- AB) × (- AB) × (- AB) > 0, then ()
A.ab0
C.a>0,b

A

It is known that a > b, 1 / A0

Because 1 / A

Given a + B = 3, ab = 0.5, find a / B + B / A

a/b+b/a
=a²/ab+b²/ab
=（a²+b²）/ab
=（a²+b²+2ab-2ab）/ab
=[（a+b）²-2ab]/ab
=（3²-2×0.5）/0.5
=（9-1）/0.5
=16

The known ab | ab | has () a.ab | 0, B.A | B | 0, C.A | 0, B | 0, D.A | 0 | B

A.ab〈0
The absolute value of ab ＜ ab | is greater than itself, indicating that it is a negative number

For any integer n, the N + 2 power of 3 - the N + 2 power of 2 + the n power of 3 - the n power of 2 must be a multiple of 10

N + 2 power of 3-N + 2 power of 2-N power of 3-N power of 2 = n power of 3 * 10-N power of 2 * 5
So it must be a multiple of 10

Solve the following inequality or system of inequalities, and express the solution set on the number axis. (1) 5 (X-2) > 4 (2x-1); (2) 2 (4x − 3) 3 ≤ 5 (5x + 12) 63x − 5 < 1 − 2x

(1) Remove the denominator, get: 5x-10 ＞ 8x-4, transfer the term, get: 5x-8x ＞ - 4 + 10, merge the similar terms, get: - 3x ＞ 6, coefficient into 1, get: X ＜ - 2.; (2) 2 (4x − 3) 3 ≤ 5 (5x + 12) 6 ①3x−5＜1−2x… ② Then the solution set of inequality system is: - 8 ≤ x < 65