(2+4+6..+2022+2024)*(125125*127-127127*125) tính nhanh
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(2+4+6+...+2017+2018)x(125125x127-127127x125)
=(2+4+6+...+2018)x(125x1001x127-127127x125)
=(2+4+6+..+2018)x(125x127127-127127x125)
=(2+4+6+...+2018)x0
=0
k mk nha!!
Chuc bn hok tot

Ta có:
(1+3+5+...+2005)x(125125x127-127127x125)
=(1+3+5+...+2005)x(125x1001x127-127x1001x125)
=(1+3+5+...+2005)x0
=0
a+b=10
a+c=25
b+c=18
a+b=10
a+c=25
b+c=18
a+b=10
a+c=25
b+c=18
a+b=10
a+c=25
b+c=18
a+b=10
a+c=25
b+c=18
a+b=10
a+c=25
b+c=18
a+b=10
a+c=25
b+c=18

( 1 + 3 + 5 +.....+ 2005) * ( 125125 * 127 - 127127 * 125 )
= ( 1+3+5+....+2005) * ( 125 * 1001 * 127 - 127 * 1001 * 125)
= (1 +3+5+....+2005) * 0
= 0
k cho mình nha bạn...!
Ta có:(1+3+5+...+2005)x(125125x127-127127x125)
=(1+3+5+...+2005)x(125x1001x127-127x1001x125)
=(1+3+5+...+2005)x0
=0

( 1 + 3 + 5 + 7 +... + 2003 + 2005 ) x ( 125125 x 127 - 127127 x 125 )
= ( 1 + 3 + 5 + 7 + ... + 2003 + 2005 ) x ( 125 x 1001 x 127 - 127 x 1001 x 125 )
= ( 1 + 3 + 5 + 7 + ... + 2003 + 2005 ) x 0
= 0
~ Thiên Mã ~
lllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooolllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllllll

Ta có :
1+3+5+7+...+2017
=> Có số số hạng là :
(2017-1):2+=1009 ( số )
Tổng của dãy : 1+3+5+7+...+2017 là :
(2017+1).1009:2=1018081
Dãy 125125 * 127 - 127127 *125 =0
=>
( 1 + 3 + 5 + ... + 2015 + 2017) * ( 125125 * 127 - 127127 *125 ) = 0
(1+3+5+7+…+2003+2005)×(125125×127-127127×125)=
(1+3+5+7+….+2003+2005)×(125×101×127- 127×101×125)
= (1+3+5+7+…+2003+2005)×0
=0

(1+3+5+....+2011)x(125125x127-127127x125)
Ta có:
=(125125x127-127127x125)
=1001x125x127-1001x127x125
=0
Vậy:(1+3+5+....+2011)x(125125x127-127127x125)=(1+3+5+....+2011)x0=0
Đề bị sai hay sao vậy, phải là : (1+3+5+7+...+2011)*(125125127-127127*125) mới đúng

Đề: Tính
(1 + 3 + 5 + ... + 2007 + 2009 + 2011) x (125125 x 127 - 127127 x 125)
= (1 + 3 + 5 + ... + 2007 + 2009 + 2011) x 0
= 0
(1+3+5+....+2007+2009+2011)×(15890875-15890875)
(1+3+5+.....+2007+2009+2011)×0
(1+3+5+.....+2007+2009+2011)

\(\left(1+3+5+7+...+2007+2009+2011\right)\left(125125\cdot127-127127\cdot125\right)\)
\(=\left(1+3+5+...+2007+2009+2011\right)\left(125\cdot127\cdot1001-127\cdot125\cdot1001\right)\)
\(=\left(1+3+5+...+2007+2009+2011\right)\cdot0\)
\(=0\)
=( 1 + 3 + 5 + ............. + 2007 + 2009 + 2011 ) x 0
= 0
HỌC TỐT
K VÀ KN NẾU CÓ THỂ

=(1+...2005)x(125x1001x127-127x1001x125)
=(1+...2005)x0
=0
Chúc bạn học giỏi

a) A = 1+4+7+...+19+22
Số số hạng là: ( 22-1) : 3 + 1= 8 ( số)
Tổng số hạng là: ( 22+1) x 8 : 2 = 92
=> A = 1+4+7...+19+22= 92
b) B = 2+5+8+...+ 26+29
Số số hạng là: (29-2) : 3 + 1= 10 ( số)
Tổng các số hạng là: ( 29 +2) x 10 : 2 = 155
=> B = 2+5+8+...+ 26+29 = 155
c) C = ( 1+3+5+...+2017+2019) * ( 125125 *127 - 127127 *125)
=C = (1+3+5+...+2017+ 2019) * ( 125125 *127 -127127 *125)
C = (1+3+5+...+2017+2019) * ( 125 * 1001 *127 - 127* 1001 *125)
C = ( 1+3+5+...+2017+2019)* 0
C = 0
1 )A = 1 + 4 + 7 + ... + 19 + 22 ( có 8 số )
A = \(\frac{\left(22+1\right)\times8}{2}\)
A = 92
2 )B = 2 + 5 + 8 + ... + 23 + 26 + 29 ( có 10 số )
B = \(\frac{\left(29+2\right)\times10}{2}\)
B =155
(2 + 4 +6 + ... + 2022 + 2024) * (125125 * 127 - 127127 * 125)
= (2 + 4 +6 + ... + 2022 + 2024) * (125 * 1001 * 127 - 127 * 1001 * 125)
= (2 + 4 +6 + ... + 2022 + 2024) * 0
= 0
=(2+4+6+...2022+2024)*(125125*127-127*1001*125)
=(2+4+6+...+2022+2024)*(125125*127-127*125125)
=(2+4+6+...+2022+2024)*0
=0