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

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

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

\(=0\)

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

 = 0 

HỌC TỐT

K VÀ KN NẾU CÓ THỂ

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~

Bn bấm vô " Câu hỏi tương tự " đi , ở đó có câu hỏi giống bn có câu trả lời cho bn tham khảo đó 

~ Hok tốt ~
#JH 

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

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

= \(\frac{1}{3}\)- \(\frac{1}{15}\)

= \(\frac{4}{15}\)

(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 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

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 ^^

=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+...+2011] x [125125 x 127-127127 x 125]

=2028096 x0

=0

Vay phep tinh tren co ket qua = 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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