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Ta có : (1+2+3+.....+100).(12+ 22 +.....+102 ).(65.111 - 13.15.37)
=A .B.(65.111-13.5.3.37)
=A.B.(65.111-65.111)
=A.B.0
=0
\(19.64+76.34\)
\(=19.64+19.4.34\)
\(=19.64+19.136\)
\(=19.\left(64+136\right)\)
\(=19.200\)
\(=3800\)
\(35.12+65.13\)
\(=5.7.12+5.13.13\)
\(=5.84+5.169\)
\(=5.\left(84+169\right)\)
\(=5.253\)
\(=1265\)
\(19991999.1998-19981998.1999\)
\(=1999.10101.1998-1998.10101.1999\)
\(=0\)
\(\left(2+4+6+......+100\right).\left(36.333-108.111\right)\)
\(=\left(2+4+6+........+100\right).\left(36.3.111-108.111\right)\)
\(=\left(2+4+6+........+100\right).\left(108.111-108.111\right)\)
\(=\left(2+4+6+.......+100\right).0\)
\(=0\)
19x64+76x34
=19x64+19x4x34
=19x(64+136)
=19x200=3800
35x12+65x13
=5x7x12+5x13x13
=5x(7x12+13x13)
=5x253=1265
19991999x1998-19981998x1999
=1999x10001x1999-1998x10001x1999
=0
( 2+4+6+... +100) x ( 36x 333 -108 x111)
=(2+4+6+...+100)x(36x111x3-36x3x111)
=(..................tự điền, như trên thôi)x0=0
(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
a, 19 x 64 + 76 x 34
= 19 x 2 x 32 + 38 x 2 x 34
= 38 x 32 + 38 x 68
= 38 x ( 32 + 68 )
= 38 x 100
= 3800
b, 35 x 12 + 65 x 13
= 35 x 12 + 65 x 12 + 65
= ( 35 + 65 ) x 12 + 65
= 100 x 12 + 65
= 1200 + 65
= 1265
d, ( 2 + 4 + 6 + ... + 1000) x ( 36 x 333 - 108 x 111 )
= ( 2 + 4 + 6 + ... + 1000 ) x ( 36 x 3 x 111 - 108 x 111)
= ( 2 + 4 + 6 + ...+ 1000 ) x ( 108 x 111 - 108 x 111 )
= ( 2 + 4 + 6 + ... + 1000 ) x 0
= 0
Câu c chịu
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 )