The third power of 5ab-9 / 2a, the second power of B-9 / 4AB + the third power of 1 / 2a, the second power of B-11 / 4ab-a, B-5, where a = 1, B = - 2

The third power of 5ab-9 / 2a, the second power of B-9 / 4AB + the third power of 1 / 2a, the second power of B-11 / 4ab-a, B-5, where a = 1, B = - 2

The solution formula is 5ab-9 / 2A ^ 3B ^ 2-9 / 4AB + 1 / 2A ^ 3B ^ 2-11 / 4ab-a ^ 3b-5 = - 4A & # 179; B & # 178; - A & # 179; B-5 = - 4 × 1 × 4-1 × 2-5 = - 16-2-5 = - 23

How about 999 and 24 / 25 times minus five

999 24 / 25 * (- 5)
=(1000-1/25)*-5
=-5000 + 1 / 5 = - 4999 and 4 / 5

Nine hundred ninety-nine and twenty-four times five

Nine hundred ninety-nine and twenty-four times five
＝（1000-1/25）x -5
=-1000x5+1/25x5
=-5000+1/5
=-4999 4 / 5