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 + 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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(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

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

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

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

\(=0\)

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

 = 0 

HỌC TỐT

K VÀ KN NẾU CÓ THỂ

(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

cảm ơn nha

Theo đề ta có:

\(\frac{\left[\left(2013-1\right):1+1\right]x\left(2013+1\right)}{2}\)x \(\left(125125x127-127x1001\right)\)= \(\frac{2013x2014}{2}\)x \(127x\left(125125-1001\right)\)= 2027091 x 127 x 124124 .

Số lớn quá,bn tự tính nha.

(1 + 3 + 5 + 7 + ...... + 2011 + 2013) x (125125 x 127 - 127127)

Gọi (1 + 3 + 5 + 7 + ...+ 2011 + 2013) là A

        (125125 x 127 - 127127) là B . Ta có:

A là dãy cách đều 2 đơn vị

Số số hạng của A là:  

(2013 - 1) : 2 + 1 = 1007 (số)

Tổng của A là: 

(1 + 2013) x 1007 : 2 = 1014049

B = 125125 x 127 - 127127 

B = 15890875 - 127127

B = 15763748 

Suy ra (1 + 3 + 5 + 7 + .......+ 2011 + 2013) x (125125 x 127 - 127127) = 1014049 x 15763748 = 1.59852129x₁₀¹³

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

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

= \(\frac{1}{3}\)- \(\frac{1}{15}\)

= \(\frac{4}{15}\)