初二数学题一道
利用乘法公式计算: (1)999^2-1002*998 (2)-19又6/7*20又1/7
(1)999²-1002*998 =999²-(999+3)(999-1) =999²-999²+999-3*999+3 =-2*999+3 =-1995. (2)-19又6/7*20又1/7 =(-139/7)(141)/7 =-[139(140+1)]/49 =-19599/49 =-399又48/49.
(1)999²-1002*998 =999²-(1000+2)(1000-2) =999²-(1000²-4) =999²-1000²+4 =-1*1999+4 =-1995. (2)-19又6/7*20又1/7 =(-139/7)(141)/7 =-[(140-1)(140+1)]/49 =-(140²-1)/49 =-19599/49 =-399又48/49.
1)999^2-1002*998 =(1000-1)^2-(1000+2)(1000-2) =1000^2-2*1000+1-(1000^2-2^2) =-2000+1+4 =-1995 2)-(19+6/7)*(20+1/7) =-(20-1/7)(20+1/7 =-[20^2-(1/7)^2] =-400+1/49 =-399-48/49 =-(399+48/49).
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