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b)Ghi đầu baì
=(1+2+3+...+100).(12+22+32+....+1002).(65.111-13.555)
=(1+2+3+...+100).(12+22+32+....+1002).(65.111-13.5.111)
=(1+2+3+...+100).(12+22+32+....+1002).(111.(65-65))
=(1+2+3+...+100).(12+22+32+....+1002).111.0
=(1+2+3+...+100).(12+22+32+....+1002).0
=0
(1^2+2^2+...+2016^2).(65.111-13.15.17)
=(1^1+2^2+...+2016^2).0
=0
a,631=631
b, 2716:910=(33)16:(32)10=348:320=328
c,(1+2+3+.....+100).(12+22+32+...+102).(65.111-13.15.17)
=(1+2+3+.....+100).(12+22+32+...+102).(13.15.17-13.15.17)
=(1+2+3+.....+100).(12+22+32+...+102).0
=0
nhớ like
(1+2+3+.....+100).(12+22+32+.....+102).(65.111-13.15.37)
=(1+2+3+.....+100).(12+22+32+.....+102)(65.111-13.555)
=(1+2+3+.....+100).(12+22+32+.....+102)(65.111-13.5.111)
=(1+2+3+.....+100).(12+22+32+.....+102)[111(65-65)]
=(1+2+3+.....+100).(12+22+32+.....+102)(100.0)
=(1+2+3+.....+100).(12+22+32+.....+102)0
=0
b) 19991999.1998-19981998.1999 = 1999.10001.1998-1998.10001.1999 = 0
c) (1+2+3+...+100).(12+22+32+...+102).(65.111-13.15.37) = (1+2+3+...+100).(12+22+32+...+102).(13.5.3.37-13.15.37)=(1+2+3+...+100).(12+22+32+...+102).0 =0
Còn a) thì mk chịu
Tính nhanh:
a) (1+2+3+..+100) x (12 +22+32+...+...102) x ( 65.111-13.15.37)
b)19x64+76x34
c)12x35+65x13
a Ta có
(1+2+3....+100).(1^2+2^2+..+10^2).(13.5.3.37-13.15.37)
=(1+2...+100).(1^2+2^2+...+10^2).0=0
b Ta co
19.64+76.34=19.4.16+76.34=76.16+76.34=76(16+34)=76.50=3100 ( cau c tuong tu )