1、(1+1/2)×(1-1/2)×···×(1+1/1999)×(1-1/1999)=?(1+1/1997+1/1998+1/1999)×(1/1997+1/1998+1/1999+1/2000)-(1+1/1997+1/1998+1/1999+1/2000)×(1/1997+1/1998+1/1999)=?各位能做多少就做多少···········第一题的答案是不

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1、(1+1/2)×(1-1/2)×···×(1+1/1999)×(1-1/1999)=?(1+1/1997+1/1998+1/1999)×(1/1997+1/1998+1/1999+1/2000)-

1、(1+1/2)×(1-1/2)×···×(1+1/1999)×(1-1/1999)=?(1+1/1997+1/1998+1/1999)×(1/1997+1/1998+1/1999+1/2000)-(1+1/1997+1/1998+1/1999+1/2000)×(1/1997+1/1998+1/1999)=?各位能做多少就做多少···········第一题的答案是不
1、(1+1/2)×(1-1/2)×···×(1+1/1999)×(1-1/1999)=?
(1+1/1997+1/1998+1/1999)×(1/1997+1/1998+1/1999+1/2000)-(1+1/1997+1/1998+1/1999+1/2000)×(1/1997+1/1998+1/1999)=?
各位能做多少就做多少···········
第一题的答案是不是1000/1999············
(1/11+1/21+1/31+1/41)×(1/21+1/31+1/41+1/51)-(1/11+1/21+1/31+1/41+1/51)×(1/21+1/31+1/41)=?

1、(1+1/2)×(1-1/2)×···×(1+1/1999)×(1-1/1999)=?(1+1/1997+1/1998+1/1999)×(1/1997+1/1998+1/1999+1/2000)-(1+1/1997+1/1998+1/1999+1/2000)×(1/1997+1/1998+1/1999)=?各位能做多少就做多少···········第一题的答案是不
1.设1/1997+1/1998+1/1999=a
(a+1)×(a+1/2000)-(1+a+1/2000)×a
=a²+a+1/2000a+1/2000-a-a²-1/2000a
=1/2000
2.设1/21+1/31+1/41=a
(1/11+a)×(a+1/51)-(1/11+a+1/51)×a
=(1/11)×(1/51)
=1/561

-2(1+2²+·····+2ⁿ)+﹙2n-1﹚·2^(n+1)=? x【x+1】分之1+【x+1】乘【x+2】分之1+······+【2x-1】乘2x分之1等于8分之1 观察下面规律:1/1×2=1-1/2;1/2×3=1/2-1/3;……(1)求和:1/1×3+1/2×3+1/3×4+······1/2009×20101/1×2+1/2×3+1/3×4+······1/2009×2010 1、(1+1/2)×(1-1/2)×···×(1+1/1999)×(1-1/1999)=?(1+1/1997+1/1998+1/1999)×(1/1997+1/1998+1/1999+1/2000)-(1+1/1997+1/1998+1/1999+1/2000)×(1/1997+1/1998+1/1999)=?各位能做多少就做多少···········第一题的答案是不 (1)5+10+15+20+25+···+85+90+95+100(2)﹙1+3+5+7+···+1991﹚-﹙2+4+6+8+···+1990﹚ 利用等差规律计算一、计算:1+3+5+···+197+99;81+79+···+13+11;1-2+3-4+5-6+···+2009-2010+2011.二、按一定规律排列的算式:4+2,5+8,6+14,7+20,···,那么第100个算式是 1+2+3+4+5········+n,=多少? 计算:1+2-3-4+5+6-7-8+···+2013+2014-2015-2016 2003/1+2003/2+2103/3+······+2003/4005 47/6×48 简便计算 42/1÷(6/1+7/1)简便计算 2/1+6/1+12/1+········· +90/1 简便计算 要 找规律(用含有N的式子表示你的发现) 2×2=1×3+15×5=4×6+1 ··· (-2)×(-2)=(-1)×(-3)+1 (-7)×(-7)=(-6)×(-8)+1···_____________________________ 1+2+3+4+···+n=? 1根据下列条件求a-b+c-d的值.a=﹣8,b=﹣12,c=7,d=42.计算:3分之1-3分之2的绝对值+4分之1-3分之一的绝对值+5分之1-4分之1的绝对值+············+10分之1-9分之1的绝对值 用简便方法计算(﹣a^m· b^n ·c)^2·(a^m-1· b^m+1· c^n)^2 解方程﹙0.5+0·2x﹚/0·03-﹙1-0·3X﹚/0·2=1 已知1+2+3+···+15=14400,求2+4+6+···+30的值 已知函数f(x)=x^2-2ax+5在区间(-∞,2]上单调递减,且对任意的x1,x2∈[1,a+1],都有|f(x1)-f(x2)|≤4,则实数a的取值范围是 ▲ .求过程啊························· 2002²-2001²+2000²-1999²+···+2²-1²