So sánh 1007/1008 và 1008/1009
So sánh: C = 1.2.3.4....2011
D = 1007/2 . 1008/2 . 1009/2 ...... 2012/2
C = 1.2.3.4....2011
D = 1007/2 . 1008/2 . 1009/2.....2012/2
D = (1007.1008.1009.....2012) : (2.2.2.2........2) (có 2012 - 1007 + 1 = 1006 số 2 )
D = (1007.1008.1009....503 .2.2) : 21004
MÀ (4.503.1007.1008.....2011) < (1.2.3.....2011)
Vậy c > D
\(\dfrac{x+1}{2014}+\dfrac{x+2}{2013}=...+\dfrac{x+1007}{1008}=\dfrac{x+1008}{1007}+\dfrac{x+1009}{1006}+...+\dfrac{x+2014}{1}\)
\(\dfrac{x+1}{2014}+\dfrac{x+2}{2013}+.....+\dfrac{x+1007}{1008}=\dfrac{x+1008}{1007}+\dfrac{x+1009}{1006}+........+\dfrac{x+2014}{1}\)\(\Leftrightarrow\left(\dfrac{x+1}{2014}+1\right)+\left(\dfrac{x+2}{2013}+1\right)+...+\left(\dfrac{x+1007}{1008}+1\right)=\left(\dfrac{x+1008}{1007}+1\right)+\left(\dfrac{x+1009}{1006}+1\right)+...+\left(\dfrac{x+2014}{1}+1\right)\)\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}+...+\dfrac{x+1007}{1008}=\dfrac{x+2015}{1007}+\dfrac{x+1009}{1006}+...+\dfrac{x+2014}{1}\)\(\Leftrightarrow\dfrac{x+2015}{2014}+\dfrac{x+2015}{2013}+...+\dfrac{x+2015}{1008}-\dfrac{x+1008}{1007}-\dfrac{x+2015}{1006}-...-\dfrac{x+2015}{1}=0\)\(\Leftrightarrow\left(x+2015\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1008}-\dfrac{1}{1007}-\dfrac{1}{1006}-...-1\right)=0\)\(\Leftrightarrow x+2015=0\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1008}-\dfrac{1}{1007}-\dfrac{1}{1006}-...-1>0\right)\)\(\Leftrightarrow x=-2015\)
Vậy x=-2015
1-1/2+1/3-1/4+1/5-1/6+...+1/2011-1/2012 / 1006-1006/1007-1007/1008-1008/1009-...-2010/2011-2011/2012
Ta có: \(\dfrac{x+1006}{1007}+\dfrac{x+1005}{1008}=\dfrac{x+1004}{1009}+\dfrac{x+1003}{1010}\)
\(\Leftrightarrow\dfrac{x+1006}{1007}+1+\dfrac{x+1005}{1008}+1=\dfrac{x+1004}{1009}+1+\dfrac{x+1003}{1010}+1\)
\(\Leftrightarrow\dfrac{x+2013}{1007}+\dfrac{x+2013}{1008}=\dfrac{x+2013}{1009}+\dfrac{x+2013}{1010}\)
\(\Leftrightarrow\dfrac{x+2013}{1007}+\dfrac{x+2013}{1008}-\dfrac{x+2013}{1009}-\dfrac{x+2013}{1010}=0\)
\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{1007}+\dfrac{1}{1008}-\dfrac{1}{1009}-\dfrac{1}{1010}\right)=0\)
mà \(\dfrac{1}{1007}+\dfrac{1}{1008}-\dfrac{1}{1009}-\dfrac{1}{1010}\ne0\)
nên x+2013=0
hay x=-2013
Vậy: S={-2013}
tính nhanh giá trị biểu thức
1-1/2+1/3-1/4+1/5-1/6+....+1/2011-1/2012
-------------------------------------------------------------
1006-1006/1007-1007/1008-1008/1009-...-2010/2011-2011/2012
---------- là phần nha
trình bày cách giải
Tính bằng cách hợp lí: (1-1/1007)*(1-1/1008)*(1-1/1009)*...*(1-1/1012)
\(\left(1-\frac{1}{1007}\right)\left(1-\frac{1}{1008}\right)\left(1-\frac{1}{1009}\right)\left(1-\frac{1}{1010}\right)\left(1-\frac{1}{1011}\right)\left(1-\frac{1}{1012}\right)\)
\(=\frac{1006}{1007}\cdot\frac{1007}{1008}\cdot\frac{1008}{1009}\cdot\frac{1009}{1010}\cdot\frac{1010}{1011}\cdot\frac{1011}{1012}\)
\(=\frac{1006\cdot1007\cdot1008\cdot1009\cdot1010\cdot1011}{1007\cdot1008\cdot1009\cdot1010\cdot1011\cdot1012}=\frac{503}{506}\)
=\(\frac{1006}{1007}.\frac{1007}{1008}.....\frac{1011}{1012}\)
=\(\frac{1006}{1012}\)
=\(\frac{503}{506}\)
nếu sai sót mong mọi người sửa lỗi đúng thì ủng hộ
\(\left(1-\frac{1}{1007}\right)\left(1-\frac{1}{1008}\right)\left(1-\frac{1}{1009}\right)...\left(1-\frac{1}{1012}\right)\)
\(=\frac{1006}{1007}\cdot\frac{1007}{1008}\cdot\frac{1008}{1009}\cdot...\cdot\frac{1011}{1012}\)
\(=\frac{1006\cdot1007\cdot1008\cdot...\cdot1011}{1007\cdot1008\cdot1009\cdot...\cdot1012}=\frac{1006}{1012}=\frac{503}{506}\)
so sánh s=1-1/2+1/3-...........+1/1013 và p=1/1007+1/1008+..........+1/2013
so sánh A và B biết ( ghi cả cách giải ra nhé)
A=1x3x5x7x9x.........x2011
B=1007 phần 2 nhân với 1008 phần 2 nhân với 1009 phần 2 nhân vân vân nhân đến 2012 phần 2
A=1-1/2-1/3-1/4-...-1/2012. B = 1/1007+1/1008+1/1009+1/2012 tính (A)/(B)^2013