-1的2次方-1\/2*【3*(-2\/3)的2次方-(-1)的2008次方】+1\/4÷ (-1\/2...
=-1-1\/2*(3*4\/9-1)+1\/4÷(-1\/8)=-1-1\/2*1\/3-2 =-3又6分之1
(1-1\/2^2)*(1-1\/3^2)*(1-1\/4^2)*...(1-1\/100)^2=???
你好!每项都用平方差公式——a^2-b^2=(a-b)(a+b),然后就简单了。原式=(1-1\/2)×(1+1\/2)×(1-1\/3)×(1+1\/3)×……×(1-1\/100)×(1+1\/100)=1\/2 × 3\/2 × 2\/3 × 4\/3 ×3\/4 ×……× 100\/99 × 99\/100 ×101\/100 =1\/2 × 101\/100 =101\/200 ...
计算:(1-1\/2)*(1-1\/3)*(1-1\/4)*……*(1-1\/10).怎么做?谢了!
原式=(1-1\/2)(1-1\/3)...(1-1\/n)=1\/2×2\/3×3\/4×...×(n-2)\/(n-1)×(n-1)\/n 可以发现,前面一个分母正好等于后面一个分子,把他们约掉,即可得 原式=1\/n。
计算:-3*(-1\/3)的3次方-(1\/3)的2次方\/(-2\/3)的2次方
-3*(-1\/3)的3次方-(1\/3)的2次方\/(-2\/3)的2次方 =-3×(-1\/27)-1\/9÷(4\/9)=1\/9-1\/4 =4\/36-9\/36 =-5\/36;请采纳 如果你认可我的回答,敬请及时采纳,~如果你认可我的回答,请及时点击【采纳为满意回答】按钮 ~~手机提问的朋友在客户端右上角评价点【满意】即可。~你...
计算:(1-1\/2^2)*(1-1\/3^2)*(1-1\/4^2)*...*(1-1\/2011^2)*(1-1\/2012^...
(1-1\/2^2)*(1-1\/3^2)*(1-1\/4^2)*...*(1-1\/2011^2)*(1-1\/2012^2)=(1-1\/2)(1+1\/2)(1-1\/3)(1+1\/3)(1-1\/4)(1+1\/4)...(1-1\/2011)(1+1\/2011)(1-1\/2012)(1+1\/2012)=1\/2*3\/2*2\/3*4\/3*3\/4*5\/4*...*2010\/2011*2012\/2011*2011\/2012*2013...
(1-1\/2×2)×(1-1\/3×3)×(1-1\/4×4)×…×(1-1\/2004×2004)=
解法为:每一项都可写成(n-1)(n+1)\/n*n 即(n-1)(n+1)\/n*n × n(n+2)\/(n+1)(n+1) ×(n+1)(n+3)\/(n+2)(n+2)……所以原式就是(3\/2*2)×(2*4\/3*3)×(3*5\/4*4)×……(2002*2004\/2003*2003)×(2003*2005\/2004*2004)=1\/2*2005\/2004=2005\/4008 看上去有点...
计算[(-3\/2)的三次方*(-4\/3)的二次方\/(-1\/2)-3的二次方-(-3)的三次方...
原式=[(-3\/2)³×(-4\/3)²÷(-1\/2)-3²-(-3)³]×(-1^4)=[(-27\/8)×(16\/9)×(-2)-9-(-27)]×(-1)=(12-9+27)×(-1)=30×(-1)=-30
...计算(-1\/2)*(-2\/3)*(1-1\/4)*(-1又1\/3)*(-1又4\/5)
(-1\/2)*(-2\/3)*(1-1\/4)*(-1又1\/3)*(-1又4\/5)=(-1\/2)*(-2\/3)*(3\/4)*(-4\/3)*(-9\/5)=[(1\/2)(2\/3)]*[(3\/4)(4\/3_]*(9\/5)=(1\/3)*1*(9\/5)=3\/5
(1-1\/2)乘以(1-1\/3)乘以(1-1\/4)乘以(1-1\/5)乘以(1-1\/6)乘以(1-1\/7...
=(1\/2)×(2\/3)×(3\/4)×(4\/5)×(5\/6)×(6\/7)前一个分母和后一个分子约分 =1\/7
(1-1\/2)乘以(1-1\/3)乘以(1-1\/4)乘以...(1-1\/9)乘以(1-1\/10
(1-1\/2)×(1-1\/3)×(1-1\/4)×...×(1-1\/10)=(1\/2)×(2\/3)×(3\/4)×...×9\/10 =1\/10