(25^2+5^11)/30=? (-2)^2001+(-2)^2002-2^2001=?
来源:百度知道 编辑:UC知道 时间:2024/05/16 11:58:30
用分解因式的方法计算
你好!
(1):
解,原式=(5^4+5^11)/(5*6) = [5^3(1+5^7)]/6 = 125*78126/6 = 1627625
(2) :
解:原式=(-2)^2001 * (1-2+1) = (-2)^2001*0 = 0
^_^
1.
(25^2+5^11)/30
=(5^4+5^11)/30
=5^4(1+5^7)/30
=5^4(1+5^7)/(5*6)
=5^3(1+5^7)/6
=5^3(1+5^7)/6
=125*(1+78125)/6
=125*(1+78125)/6
=9765750/6
=1627625
2.
(-2)^2001+(-2)^2002-2^2001
=-2^2001+2^2002-2^2001
=-2^2001(1-2^1+1)
=-2^2001(1-2+1)
=0
(25^2+5^11)/30
=(25^2+25^2*5^7)/(5*6)
=5^4(1+5^7)/(5*6)
=5^3(1+5^7)/6
=125*78126/6
=1627625
(-2)^2001+(-2)^2002-2^2001
=-2^2001+2^2002-2^2001
=-2^2001*(1-2+1)
=-2^2001*0
=0
第一题:(25^2+5^11)/30=[(5^2)^2+5^11]/30
=(5^4+5^11)/30
=5^4(1+5^7) /30
=5^4*78126/30=1627625
第二题:(-2)^2001+(
11/2*11/3*11/4*11/5*11/6*11/7*11/8
因式分解a2b2+1/25c2+2/5abc
2/3+2/(3×5)+2/(5×7)+2/(7×9)+2/(9×11)+2/(11×13)
1/(2+5)+1/(5+8)+1/(8+11)+1/(11+14)+...+1/(47+50)
2/2+3/4+4/8+5/16+---+11/1024
数列3/5,1/2,5/11,3/7,7/11,......通项公式
1/2+5/6+11/12+19/20+29/30=
找规律:1/2 2/5 3/10 4/17 25/26 -------
1/3 2/5 4/7 ( ) 256/11
1/2+5/6+11/12+...+109/110等于多少?