It is known that | a + & # 189; | + (B-3) √ 2 = 0, find [(2a + b) √ 2 + (b-2a) (2a-b) - 6B] △ 2B

It is known that | a + & # 189; | + (B-3) √ 2 = 0, find [(2a + b) √ 2 + (b-2a) (2a-b) - 6B] △ 2B

∵|a+12|+（b-3）2=0,
A + 12 = 0 or B-3 = 0,
The solution is a = - 12, B = 3;
∵【（2a+b）2+（2a+b）（b-2a）-6b】÷2b
=（4a2+4ab+b2+b2-4a2-6b）÷2b
=2b（b+2a-3）÷2b
=b+2a-3,
The original formula is 3 + 2 × (- 12) - 3 = - 1;
So the answer is: - 1

Why is 2 (a + b) (a-b) equal to the square of 2A - the square of 2B? Why is it not the square of 2A - the square of B

2(a+b)(a-b)
=2（a²-b²）
=2a²-2b²

If 1a − 1b = 4, then the value of a − 2Ab − B2A − 2B + 7ab is equal to ()
A. 6B. -6C. 215D. −27

It is known that 1a − 1b = 4 can give A-B = - 4AB, then a − 2Ab − B2A − 2B + 7ab = a − B − 2ab2 (a − b) + 7ab = − 4AB − 2Ab − 8ab + 7ab = − 6ab − AB = 6

If 1 / A + 1 / b = 4, then a + 2Ab + B / 2A + 7ab + 2b is equal to

(a+2ab+b)/(2a+7ab+2b)
Divide the numerator and denominator by ab at the same time
Original formula = (1 / B + 2 + 1 / a) / (2 / B + 7 + 2 / a)
=(4+2)/(2*4+7)
=6/15
=2/5