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

A= 1 + 4 + 7 +...+ 19 + 22

  = [ ( 22 - 1 ) : 3 + 1 ] x ( 1 + 22 ) : 2

 = [ 21 : 3 + 1 ] x 23 : 2

 = [ 7 + 1 ] x 23 : 2

 = 8 x 23 : 2

 = 4 x 23 = 92

B = 2 + 5 + 8 + ... + 26 + 29

   = [(29 - 2 ) : 3 + 1] x ( 2 + 29 ) : 2

   = [27 : 3 + 1] x 31 : 2

   = [9 + 1] x 31 : 2

   = 10 x 31 : 2 = 310 : 2 = 155

C = ( 1 + 3 + 5 + ... + 2017 + 2019 ) x ( 125125 x 127 - 127127 x 125 )

  = ( 1 + 3 + 5 + ... + 2017 + 2019 ) x ( 125 x 1001 x 27 - 127 x 1001 x 125 )

  = ( 1 + 3 + 5 + ... + 2017 + 2019 ) x 0 = 0

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

thank you VRCT _gnk_Thùy Linh

( 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 + ...+ 2005 )  . ( 125125 . 127 - 127127 . 125 )

= ( 1 + 3 + 5 + ... + 2005 ) .  ( 125 . 1001 x 127 - 127 x 1001 x 125 )

= ( 1 + 3 + 5 + .. + 2005 ) .          0

=   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

Xin lỗi

(1+3+5+…+2015+2017)×(125125×127-127127×125)=
(1+3+5+….+2015+2017)×(125×101×127- 127×101×125)
= (1+3+5+…+2015+2017)×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

a. 

(1 + 3 + 5 + ... + 2007 + 2009 + 2011) x (125125 x 127 - 127127 x 125)

= (1 + 3 + 5 + ... + 2007 + 2009 + 2011) x 0

= 0

b.

\(\frac{2006\times125+1000}{126\times2006-1006}=\frac{2006\times125+1000}{125\times2006+1\times2006-1006}=\frac{2006\times125+1000}{125\times2006+1000}=1\)

a, 

( 1+3+5+7+…+2003+2005) x (125 125 x 127 – 127 127 x 125)

= ( 1+3+5+7+…+2003+2005) x (125 x 1001 x 127 – 127 x 1001x 125)

=  ( 1+3+5+7+…+2003+2005) x                   0  =    0

A = 0

B= 2006

C= 1993

Tui làm sai á ^^

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)

(1 + 3 + 5 + .... +2007 + 2009 + 2011 ) x (125125 x 127 - 127127 x 125 )

= ( 1 + 3 + 5 +..................+ 2007 + 2009 + 2011 ) x ( 15890875 - 15890875 )

= ( 1 + 3 + 5 +...................+ 2007 + 2009 + 2011 ) x 0

= 0

Chúc bạn học tốt!!!

\(\left(1+3+5+7+...+2007+2009+2011\right)\left(125125\cdot127-127127\cdot125\right)\)

\(=\left(1+3+5+...+2007+2009+2011\right)\left(125\cdot127\cdot1001-127\cdot125\cdot1001\right)\)

\(=\left(1+3+5+...+2007+2009+2011\right)\cdot0\)

\(=0\)

 =( 1 + 3 + 5 + ............. + 2007 + 2009 + 2011 ) x 0 

 = 0 

HỌC TỐT

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