Given a, B ∈ R, prove 2 (A2 + B2) ≥ (a + b) 2

Given a, B ∈ R, prove 2 (A2 + B2) ≥ (a + b) 2

(12 points in this question) prove: to prove: 2 (A2 + B2) ≥ (a + b) 2 as long as the proof 2A2 + 2B2 ≥ A2 + B2 + 2Ab as long as the proof A2 + B2 ≥ 2Ab & nbsp; & nbsp; (5 points) that is (a-b) 2 ≥ 0, and this formula is obviously true, so 2 (A2 + B2) ≥ (a + b) 2 holds (12 points)

If|is equal to 2 + | - 1245?

/ A + 1 / 2 / + / B-5 / 3 / = 0 and the absolute value is nonnegative, a + 1 / 2 = 0, B-5 / 3 = 0, A-1 / 2, B = 5 / 3

What is the sum of three powers of formula a plus three powers of formula B?

By using the formula of cubic sum (difference) in multiplication formula in reverse, the formula of cubic sum (difference) in factorization is obtained
a3+b3=(a+b)(a2-ab+b2)
a3-b3=(a-b) (a2+ab+b2)

Given that a, B and C are three sides of any triangle, try to judge whether a ^ 2-2ab + B ^ 2-C ^ 2 is greater than zero, less than zero or equal to zero?

Original formula = (a-b) ^ 2-C ^ 2 = (a-b-c) (a-b + C)
A B C is that the three sides (a-b-c) of the triangle are less than 0 (a-b + C) and greater than 0, so the original formula is less than 0

If | ab-3 | + B-1 | = 0, try to find AB / 1 + (a + 1) (B + 1) / 1 + (a + 2) (B + 2) / 1 +(a + 2012 (b2012) / 1

If | ab-3 | + B-1 | = 0,
ab-3=0 ,b-1=0
Solution
a=3,b=1
ab/1+(a+1)(b+1)/1+(a+2)(b+2)/1… +(a + 2012 (b2012) / 1
=(3*1)/1+(3+1)(1+1)/1+(3+2)(1+2)/1… +(3+2012)(1+2012)/1
=2/1 *[(3*1)/2+(4*2)/2+(5*3)/2… +(2015*2013)/2]
=2/1 *[1-3/1+2/1 -4/1+3/1-5/1… +2013/1-2015/1]
=2/1 *[1+2/1 -2014/1-2015/1]
=2/1 *[2014/3021 -2014/1-2015/1]
=2/1 *[1007/1510-2015/1]
=2/1 *[(1007*2015)/(1510*2015-1)]
=4058210/3042649

|A + B-7 | + (AB + 3) = 0 to find the value of 5a-3ab-4b-7a-11ab + 2B

Observe the structure a2＋b2＝7… AB = - 3 is substituted into the following formula solve

If | A-3 | + | B + 1 | 0, then AB?

|a-3|+|b+1|=0
So A-3 = 0
b+1=0
a=3,b=-1
So AB = - 3

If a > 0, b > 0. A ^ 3 + B ^ 3 ≥ a ^ 2B + AB ^ 2
Such as the title

The inverse method assumes that a ^ 3 + B ^ 3 ≥ a ^ 2B + AB ^ 2
Then a ^ 3-A ^ 2B + B ^ 3-AB ^ 2 ≥ 0
Finally, it is reduced to (a + b) (a-b) ^ 2 ≥ 0
Because a > 0, b > 0, so a + b > 0, (a-b) ^ 2 ≥ 0
So the original formula holds

Given that a and B are integers and satisfy a (√ 2 + 1) + 3 (B-2 √ 2) = 6 + 3 √ 2, find the value of a + B

√2 a＋a＋3b－6√2＝6＋3√2
a＋3b＋a·√2＝6＋9√2
A = 9
a＋3b＝6 9＋3b＝6 b＝－1
∴a＋b＝9＋（－1）＝8

For two integers a, B, there are a@b Equal to (a + b) a,a@b Equal to a &; B plus 1, find [(- 2) @ (- 5)] @ (- 4)

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