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b. (2+4+6+...+100).(36.333.108.111)
=(2+4+6+...+100). 0
= 0
(12+22+32+...+102).(1+3+5...+97+99).(36.333-108.111) [ Đặt (12+22+32+...+102).(1+3+5...+97+99) = A]
= A. (36.3.111-108.111)
= A.(108.111-108.111)
= A.0
= 0
Vậy tích trên bằng 0
(12+22+32+...+102).(1+3+5...+97+99).(36.333-108.111)
= ( 12 + 22 + 32 + ... + 102 ) . ( 1 + 3 +5 + ... + 97 + 99 ) . ( 36 . 3 . 111 - 36 . 3 . 111 )
= ( 12 + 22 + 32 + ... + 102 ) . ( 1 + 3 +5 + ... + 97 + 99 ) . 0
= 0
Đáp án cần chọn là: A
(2+4+6+...+100)(36.333−108.111)
=(2+4+6+...+100)(36.3.111−36.3.111)
=(2+4+6+...+100).0=0
1.(5.3^11+4.3^12):(3^9.5^2-3^9.2^3)
=(5.3^11+4.3^12):(3^9.5^2-3^9.2^3)
=(5.3^11+4.3^11.3):[3^9.(5^2-2^3)]
=(5.3^11+12.3^11):[3^9.17]
=3^11.(5+12):(3^9.17)
=(3^11.17):(3^9.17)
=17.(3^11:3^9)
=17.3^2
=17.9
=153
2.(12+22+32+...+992+1002).(36.333-108.111)
=2.(12+22+32+...+992+1002).(36.333-36.3.111)
=2.(12+22+32+...+992+1002).(36.333-36.333)
=2.(12+22+32+...+992+1002).0
=0
a) (2+4+6+.....+100) . (36.333-108.111)
= (2+4+6+...+100).(36.111.3-108.111)
= (2+4+6+...+100).(108.111-108.111)
= (2+4+6+...+100).0
=0
b) 19991999.1998-19981998.1999
= 1999.10001.1998-1998.10001.1999
= 0
a) ( 2 + 4 + 6 + ... + 100 ) . ( 36.333 - 108.111 )
= ( 2 + 4 + 6 + ... + 100 ) . ( 36.3.111 - 108.111)
= ( 2 + 4 + 6 + ... + 100 ) . ( 108.111 - 108.111 )
= ( 2 + 4 + 6 +... + 100 ) . 0
= 0
b) 19991999.1998 - 19981998 . 1999
= 1999.10001.1998 - 1998.10001 . 1999
= 0
(2+4+6+...+100).(36.333-108.111
=(2+4+6+...+1000).(36.3.111-108.111)
=(2+4+6+...+1000.(108.111-108.111)
=(2+4+6+...+100).0
0
(1+2+3+...+1000).(108.111-36.333 )
=(1+2+3+...+1000).(108.111-36.3.111 )
=(1+2+3+...+1000).(108.111-108.111)
=(1+2+3+...+1000).0
= 0
2 ) Tính
(1+3+5+....+2017 ) . (108 . 111 - 36.333)
= (1+3+5+...+2017).(108.111-36.111.3)
=(1+3+5+....+2017).[108.111-(36.3).111]
=(1+3+5+....+2017).(108.111 - 108.111]
=(1+3+5+...+2017). 0 = 0