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

K VÀ KN NẾU CÓ THỂ

a) ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 ) 

vì  ( 125125 x 127 - 127127 x 125 ) =[125125 x (125+2)] - 127127 x 125 ) =>125125 x (125+2)=125.125125+125125.2=125125.125+250250=125125.125+125.2002=125.(125125+2002)=125.127127

=> ( 125125 x 127 - 127127 x 125 )=127127.125-127127.125=0

=>  (1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 ) =0

a) ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 ) 

= ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 )  x 0

= 0

b, \(\frac{1}{3}\)+ \(\frac{1}{15}\)+ \(\frac{1}{35}\)+ \(\frac{1}{63}\)+ \(\frac{1}{99}\)+ \(\frac{1}{143}\)+ \(\frac{1}{195}\)

= \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+\(\frac{1}{7}\)- \(\frac{1}{9}\)+...........+\(\frac{1}{13}\)- \(\frac{1}{15}\)

= \(\frac{1}{3}\)- \(\frac{1}{15}\)

= \(\frac{4}{15}\)

Đề bài có lẽ sai thừa số thứ hai phải là 125125x127-127127x125

Khi đó thừa số thứ hai sẽ là

125x1001x127-127x1001x125=0

Tích trên sẽ bằng 0
 

=0 vì thừa số sau bằng 0 nên tích bằng 0

=(1+...2005)x(125x1001x127-127x1001x125)

=(1+...2005)x0(cả hai vế giống nhau nên trừ đi thì =0)

=0

=0

nếu muốn có cách giải thì phải k 2 phát

=(1+...2005)x(125x1001x127-127x1001x125)

=(1+...2005)x0

=0

Chúc bạn học giỏi 

( 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+...+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

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