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標題:

F.2 math easy identity plz help

發問:

Using identities to find the values of the following: 1) 81^2-19^2 2) 275^2-225^2 3) 345^2-344^2 4) 538^2-462^2 5) 1001^2 6) 999^2

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1) 81^2-19^2 =(81+19)(81-19) =100x62 =6200 2) 275^2-225^2 =(275+225)(275-225) =500x50 =25000 3) 345^2-344^2 =(345+344)(345-344) =689x1 =689 4) 538^2-462^2 =(538+462)(538-462) =1000x76 =76000 5) 1001^2 =(1000+1)^2 =1000^2+2x1000x1+1^2 =1000000+2000+1 =1002001 6) 999^2 =(1000-1)^2 =1000^2-2x1000x1+1^2 =1000000-2000+1 =998001

其他解答:

1) 81^2-19^2 =(81+19)(81-19) =100x62 =6200 2) 275^2-225^2 =(275+225)(275-255) =500x20 =10000 3) 345^2-344^2 =(345+344)(345-344) =689x1 =689 4) 538^2-462^2 =(538+426)(538-426) =964x112 =107968 5) 1001^2 =1001x1001 =1002001 6) 999^2 =999x999 =998001|||||1) 81^2 - 19^2 = (81+19)x(81-19) = 100x62 = 6200 2) 275^2 - 225^2 = (275+225)x(275-225) = 500x50 = 25000 3) 345^2 - 344^2 = (345+344)x(345-344) = 689x1 = 689 4) 538^2 - 462^2 = (538+462)x(538-462) = 1000x76 = 76000 5) 1001^2 = (1000+1)^2 = 1000^2 + 2x1000x1 + 1^2 = 1000000 + 2000 + 1 = 1002001 6) 999^2 = (1000-1)^2 = 1000^2 - 2x1000x1 + 1^2 = 1000000 - 2000 + 1 = 998001 RULES : a^2 - b^2 = (a+b)x(a-b) (a+b)^2 = a^2 + 2ab + b^2 (a-b)^2 = a^2 - 2ab + b^2A9A3995907B431A4