(1+3+5+........+2011) x (125125 x 127-127127 x 125)

=(1+3+5+............+2011) x (125 x 1001 x 127-127 x 1001 x 125)

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

=0

[ 1 + 3 + 5 + 7 + ... + 2011 + 2013 ] x [ 125125 x 127 - 127127 x 125 ]

= [ ( 2013 + 1 ) x 1007 : 2 ] x [ ( 125 x 1001 )  x 127 - ( 127 x 1001 ) x 125 ]

= 1008007 x [ 125 x 1001 x 127 - 127 x 1001 x 125 ]

= 1008007 x [ ( 125 - 125 ) x ( 127 - 127 ) x ( 1001 - 1001 ) ]

= 1008007 x [ 0 x 0 x 0 ]

= 1008007 x 0

= 0

= minh chiu ban tu lam

Ta có : ( 1+3+5+7+...+2011+2013 ) x ( 125125x127-127127x125 )

=         ( 1+3+5+7+...+2011+2013 ) x ( 125x1001x127-127x1001x125 )

=         ( 1+3+5+7+...+2011+2013 ) x 0

=         0

Tk và kb vs mk nak ^_^

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

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

=0

Mình đồng ý với bạn Trần Thị Yến Vi

 ( (1+3+5+7+.......+2003+2005)x125125x127-127127x125)

= (1+3+5+7+.......+2003+2005)x 0 

= 0

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

\(=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{13\cdot15}\)

\(=\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{13\cdot15}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{15}\right)\)

\(=\frac{1}{2}\cdot\frac{14}{15}\)

\(=\frac{7}{15}\)

Sửa đề chút nhé:

\(\left(1+3+5+7+...+2009+2011\right).\left(125125.127-127127.125\right)\)

\(=\left(1+3+5+7+...+2009+2011\right).\left(125.1001.127-127.1001.125\right)\)

\(=\left(1+3+5+7+...+2009+2011\right).0\)

\(=0\)

Ý b tham khảo bài bạn nguyen thi thuy linh nhé

\(\text{Tính nhanh : }\)

\(a,\text{ }1+3+5+7+9+\text{...}+2007+2009+2011\cdot\left(125125\cdot127+127127\cdot125\right)\)

\(=\left\{\left(2009-1\right)\text{ : }2+1\right\}\cdot\left(2009+1\right)\text{ : }2+2011\cdot\left(125125\cdot127+127127\cdot125\right)\)

\(=1005\cdot2010\text{ : }2+2011\cdot\left(125125\cdot127+127127\cdot125\right)\)

\(=2020050\text{ : }2+2011\cdot\left(125125\cdot127+127127\cdot125\right)\)

\(=1010025+2011\cdot\left(125125\cdot127+127127\cdot125\right)\)

\(=1010025+2011\cdot\left(15890875+15890875\right)\)

\(=1010025+2011\cdot15890875\cdot2\)

\(=1010025+31956549625\cdot2\)

\(=1010025+63913099250\)

\(=63914109275\)

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

\(=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}+\frac{1}{13\cdot15}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{13}-\frac{1}{15}\)

\(=1-\frac{1}{15}\)

\(=\frac{14}{15}\)

=(1-2-3+4)+(5-6-7+8)+...+(2005-2006-2007+2008)+2009

=2009

1 ) 

= (1 + 3+ 5+ .....+2003+2005) \(\times\)( 125 nhân 1001 NHÂN 127 - 127 nhân 1001 nhân 125 )

= (1 + 3+ 5+ .....+2003+2005) \(\times\)0

= 0

Chúc bạn học tốt

Trả lời:

Bài 1 

\(\left(1+3+5+...+2003+2005\right)\times\left(125125\times127-127127\times125\right)\)

\(=\left\{\left(2005+1\right)\times\left[\left(2005-1\right)\div2+1\right]\div2\right\}\times\left(125\times1001\times127-127\times1001\times125\right)\)

\(=\left(2006\times1003\div2\right)\times0\)

\(=10061009\times0\)

\(=0\)

Bài 2 

\(y-6\div2-\left(48-24\times2\div6-3\right)=0\)

\(y-3-\left(48-8-3\right)=0\)

\(y-3-37=0\)

\(y-40=0\)

\(y=40\)

Vậy \(y=40\)

2) 

y - 3 - ( 48 - 48 : 6 - 3 ) = 0

y - 3 - ( 48 - 8 - 3 ) = 0

y - 3 - 32 = 0

y - 3 = 0 + 32

y - 3 = 32

y = 32+3

y = 35

Chúc bạn học tốt nhoa ! ~

1) 1-3+5-7+...-2007+2009-2011

= ( 1-3 ) + ( 5-7) + ... + ( 2009-2011 )

= -2 + (-2) + (-2) +... + (-2)

= -2 . 1000

= -2000

2) Mình chịu , não của mình chỉ được như vậy thôi

HỌC TỐT !

Bài 1:Tính các tổng đại số sau

1-3+5-7+...-2007+2009-2011

Bài 2:Tính nhanh

-69+53+46+(-94)+(-14)+78

1. 1 - 3 + 5 - 7 + ... - 2007 + 2009 - 2011

= ( 1 - 3 ) + ( 5 - 7 ) + ... + ( 2009 - 2011 )

= -2 . ( -2 ) . ... . ( -2 )

= -2 . 503

= -1006

2. -69 + 53 + 46 + ( -94 ) + ( -14 ) + 78

= [ -69 + ( -94 ) + ( -14 ) + ( 53 + 46 + 78 )

= -177 + 177

= 0

a: \(=\dfrac{4}{5}\cdot\dfrac{7}{3}\cdot\dfrac{3}{7}\cdot\dfrac{5}{4}+2011=2011+1=2012\)

b: \(=\dfrac{1}{3}\cdot\dfrac{9}{7}\cdot\dfrac{7}{9}=\dfrac{1}{3}\)

`a)4/5:3/7xx3/7:4/5+2011`

`=4/5xx7/3xx3/7xx5/4+2011`

`=(4/5xx5/4)xx(7/3xx3/7)+2011`

`=1xx1+2011`

`=1+2011`

`=2012`

__

`b)1/2xx2/3:7/9xx7/9`

`=1/2xx2/3xx9/7xx7/9`

`=(1/2xx2/3)xx(9/7xx7/9)`

`=1/3xx1`

`=1/3`