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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
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
( 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
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
\(\left(1+3+5+7+...+2007+2009+2011\right)\left(125125\cdot127-127127\cdot125\right)\)
\(=\left(1+3+5+...+2007+2009+2011\right)\left(125\cdot127\cdot1001-127\cdot125\cdot1001\right)\)
\(=\left(1+3+5+...+2007+2009+2011\right)\cdot0\)
\(=0\)
=( 1 + 3 + 5 + ............. + 2007 + 2009 + 2011 ) x 0
= 0
HỌC TỐT
K VÀ KN NẾU CÓ THỂ
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
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
(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