Calculated using the squared difference formula

1) Original formula = [(99 -1)x99] 98=[(99+1)(99-1)x99] 98=9900

2) Original formula = 2 (1-2011) (2010x2012) = 2 [(1-2011) (1 + 2011)] [(2011-1) (2011 + 1)] = -2

1) Original = [(4x +25)(2x+5)(2x-5)] [(4x +25)(2x+5)]=2x-5

2) Original formula = [y(x+3)(x-3)] [y(x+3)] = x-3

1) Original formula = [(m+2n)+(2m-n)][m+2n)-(2m-n)]=(3m+n)(3n-m)=3n +5mn-3m

99 to the 3rd power - 99) 98 = 99 * (99 2-1)) 98 = [99 * (99 + 1) (99-1)] 98

2010 1/1006-2011 to the power of 2 = (1005 + 1006) 1006-2011 2 = 1005 1006 + 1-2011 2

16x to the power of 4 - 625) (2x + 5) (4x squared + 25) = (16x 4-25 2) (2x + 5) (4x 2 + 25).

4x^2+25)(4x^2-25)÷(2x+5)÷(4x^2+25)

4x^2+25)(2x-5)(2x+5)]÷2x+5)÷(4x^2+25)

2x-5)x to the power of 2 y-9y) (xy+3y)=y(x 2-9) y(x+3)=[y(x+3)(x-3)] y(x+3)=x-3

9 (m+2n) squared - (2m-n) squared = [3(m+2n)-(2m-n)][3(m+2n)+(2m-n)].

5（m+7n）*（m+n）

99 to the 3rd power - 99) 98 = 99 * (99 2-1)) 98 = [99 * (99 + 1) (99-1)] 98

2010 1/1006-2011 to the power of 2 = (1005 + 1006) 1006-2011 2 = 1005 1006 + 1-2011 2

16x to the power of 4 - 625) (2x + 5) (4x squared + 25) = (16x 4-25 2) (2x + 5) (4x 2 + 25).

4x^2+25)(4x^2-25)÷(2x+5)÷(4x^2+25)

2x-5)

This problem is easier to calculate with a perfectly squared formula:

Original = (3x-4y) -3x+y).

9x²-24xy+16y²-（9x²+6xy+y²）=9x²-24xy+16y²-9x²-6xy-y²=-30xy+15y²

(1) (4/9x²-y²)

3)4a²-9b²

Hope it helps.

Learning progress o ( o thank you.