有规律数学题(1-2²/1)×(1-3²/1)×(1-4²/1).(1-2009²
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有规律数学题
(1-2²/1)×(1-3²/1)×(1-4²/1).(1-2009²/1)×(1-2010²/1)是多少
252²-248²/100²是多少
我写错了
(1-2²/1)×(1-3²/1)×(1-4²/1).(1-2009²/1)×(1-2010²/1)是多少
252²-248²/100²是多少
我写错了
你好像分子、分母都写反了啊.
(1-1/2^2)(1-1/3^2)...(1-1/2010^2)
=(1+1/2)(1-1/2)(1+1/3)(1-1/3)...(1+1/2010)(1-1/2010)
=(1/2)(3/2)(2/3)(4/3)(3/4)(5/4)...(2008/2009)(2010/2009)(2009/2010)(2011/2010)
=(1/2)(2011/2010)
=2011/4020
100^2/(252^2-248^2)
=100^2/[(252+248)(252-248)]
=100^2/(500×4)
=100×100/(100×20)
=100/20
=5
如果是(252^2-248^2)/100^2
(252^2-248^2)/100^2
=[(252+248)(252-248)]/100^2
=(500×4)/100^2
=100×20/(100×100)
=20/100
=1/5
(1-1/2^2)(1-1/3^2)...(1-1/2010^2)
=(1+1/2)(1-1/2)(1+1/3)(1-1/3)...(1+1/2010)(1-1/2010)
=(1/2)(3/2)(2/3)(4/3)(3/4)(5/4)...(2008/2009)(2010/2009)(2009/2010)(2011/2010)
=(1/2)(2011/2010)
=2011/4020
100^2/(252^2-248^2)
=100^2/[(252+248)(252-248)]
=100^2/(500×4)
=100×100/(100×20)
=100/20
=5
如果是(252^2-248^2)/100^2
(252^2-248^2)/100^2
=[(252+248)(252-248)]/100^2
=(500×4)/100^2
=100×20/(100×100)
=20/100
=1/5
(2²+4²+6²+.+98²+100²)-(1²+3&su
有规律数学题(1-2²/1)×(1-3²/1)×(1-4²/1).(1-2009²
1²-2²+3²-4²+5²-6²+…-100²+
4²+3²>2*4*3,(-2)²+1²>2*(-2)*1,2²+2&
100²-99²-98²-97²-.-1²
1²/(1²-100+5000)+2²/(2²-200+5000)+.k&sup
1.计算:1²+4²+6²+7²=102,2²+3²+5&s
不用计算器求值1²+2²+3²+4²+5²+6²+7&sup
(1-1/2²)(1-1/3²)(1-1/4²).(1-1/10²)=?
2(x²-1²)
100²-99²+98²-97²+96²+……+2²-1&s
100²-99²+98²-97²+...+2²-1²怎么因式