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

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

( 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 ~

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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

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

1. (1 + 3 + 5 + ... + 2005) . ( 125125.127 - 127127.125)

Ta có 125125.127 - 127127.125

=125.1001.127 - 127.1001.125

=0

=> (1 + 3 + 5 + ... + 2005) . ( 125125.127 - 127127.125)

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

=0

Bài 1:

\(\left(1+3+5+...+2005\right)\cdot\left(125125.127-127127.125\right).\)

\(=\left(1+3+5+...+2005\right)\left(125.1001.127-127127.125\right).\)

\(=\left(1+3+5+...+2005\right)\left(125.127127-127127.125\right).\)

\(=\left(1+3+5+...+2005\right).0=0.\)

Vậy.....

Bài 2:

\(\left(7.13+8.13\right):\left(\dfrac{29}{3}-x\right)=39.\)

\(\left[\left(7+8\right)13\right]:\left(\dfrac{29}{3}-x\right)=39.\)

\(\left[15.13\right]:\left(\dfrac{29}{3}-x\right)=39.\)

\(195:\left(\dfrac{29}{3}-x\right)=39.\)

\(\dfrac{29}{3}-x=195:39.\)

\(\dfrac{29}{3}-x=5.\)

\(\Rightarrow x=\dfrac{29}{3}-5.\)

\(\Rightarrow x=\dfrac{29}{3}-\dfrac{15}{3}=\dfrac{14}{3}.\)

Vậy.....

~ Hok tốt!!! ~

Câu hỏi của Nguyễn Đình Dũng - Toán lớp 5 - Học toán với OnlineMath

https://olm.vn/hoi-dap/detail/5547661103.html

Câu hỏi tại link này 

Câu hỏi của Nguyễn Đinh Dũng - toán lớp 5 -Học toán với OnlineMath 

https://olm.vn/hoi-dap/detail/5547661103

~Hok tốt~

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