Therefore, M=S(T+A2003)=S×T+A2003×S, n = (s+a2003) t = s× t+a2003× t.
Because s = t+a1>; T, so M> is ordinary.
Second, because 3×6= 18, 2a×2b=2c.
That is, c=2ab.
( 1) 1.(x+7)(x+9)= x2+ 16x+63;
2.(x- 10)(x+20)=x^2+ 10x-200;
3.(x-3)(x-2)=x^2-5x+6
(I wonder if the title means this)
(2).(x+a)(x+b)=x^2+(a+b)x+ab
(3).(t+ 1\6)(t- 1\7)=t^2+ 1/42t- 1/42
(4) Because ab=36 and A and B are integers.
So ab =1× 36 = 2×18 = 3×12 = 4× 9 = 6× 6.
So m=a+b= 1+36 or 2+ 18 or 3+ 12 or 4+9 or 6+6.
M=37 or 20 or 15 or 13 or 12.
(1) Original number:11x-10; New number: 1 1x- 1.
(2)from( 1 1x- 1)( 1 1x- 10)=( 1 1x- 10)2+405。
Solution: x=5
The original number was 45.