Math Problem Know that the answer is 2007 Finding the process or explanation Thank you

Solution: (a+c) symbol b=n+c

A symbol (b+c) = n-2c

1+c) symbol: 1=n+c=2

1 symbol (1+c) = n-2c = 2

n+c=2 n-2c=2

n=2 c=0

a+b=n2010 symbol: 2010=2010+2010=4020 looks.

(a+c) operator b=n+c 1 operator 1=2 (1+c) operator 1=2+c

1+2009) operator 1=2+2009=2011 and a operator (b+c) = n-2c

1+2009) operator (1+2009) = 2011-2 * 20092010 operator 2010 = -2007

From the meaning of the title: c@b=c a@c=-2c (a+c) operator b=n+c 1 operator 1=2

1+c) operator 1=2+c

1+2009) operator 1=2+2009=2011 and a operator (b+c) = n-2c

1+2009) operator (1+2009) = 2011-2 * 20092010 operator 2010 = -2007

Only one. In a group of 4, the first defeat is slowly added to the two large ones, and then the two small ones are subtracted.

Therefore, 2012 + 2011-2010-2009 = 4 = 4 + 3-2-1 has a total of 2012 Chamo 4 = 503 groups.

Therefore, it is equal to pin resistance 503 * 4 = 2012

1 1 (2006 and 2006 2007) = 1 [(2006*2007+2006) 2007]=2007 (2006*2008).

Original = 2007 Sun Lazhao 2008 + 1 Bad 2008 = 1

Solution: Original formula = (2010 + 1) * (2010-1) - (2010 + 2) * (2010-2).

3. Use the square difference formula: (a+b)(a-b)=a2-b2

2): = 7/9 ((1/1- 1/10) +1/10-1/19) +1/19- 1/28) +1/82- 1/91))

Note: 1 1*10 = (1- 1 10) *1 9 Other terms and so on.

1 (2011+1)*39/2011=1+1/39

2 If you don't understand the second question, you should add the ** brackets of the brackets.

Or 2007 and 2008/2007 is actually (2007*2008+2007) 2008=2007*2009 2008

1.Original = (2008 7 -3-2-1+1+2+3) 2008

7 2.The product of the two formulas is equal, (20082008+10001)*2008=20082008*(2008+1) 10001*2008=20082008 20082008-20082008=0 3 9999×2222+3333×3334 ＝3333×3×2222+3333×3334 ＝3333×6666+3333×3334 ＝3333×（6666+3334） ＝3333×10000 ＝33330000