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7 tháng 3 2023
1, \(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\) ( Trừ mỗi vế cho 2 ta được phương trình như này nhé ! )
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
Do \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\) nên \(x-2010=0\Leftrightarrow x=2010\)
2, \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{55-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)+\left(\dfrac{51-x}{49}+1\right)=0\)
\(\Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\) \(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\) Do \(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\ne0\) nên \(100-x=0\Leftrightarrow x=100\)
23 tháng 7 2023
\(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
\(\dfrac{x-1}{2009}-1+\dfrac{x-2}{2008}-1=\dfrac{x-3}{2007}-1+\dfrac{x-4}{2006}\)
\(\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
\(\left(x-2010\right)\times\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
Vì \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\)
=> \(x-2010=0\)
\(x=2010\)
23 tháng 7 2023
\(\dfrac{x-1}{2009}\)+\(\dfrac{x-2}{2008}\)=\(\dfrac{x-3}{2007}\)+\(\dfrac{x-4}{2006}\)
=>\(\dfrac{x-1}{2009}\)-1+\(\dfrac{x-2}{2008}\)+1=\(\dfrac{x-3}{2007}\)-1+\(\dfrac{x-4}{2006}\)-1
=>(x-2010)x(\(\dfrac{1}{2009}\)+\(\dfrac{1}{2008}\)-\(\dfrac{1}{2007}\)-\(\dfrac{1}{2006}\))=0
=>x-2010=0 (vì \(\dfrac{1}{2009}\)+\(\dfrac{1}{2008}\)-\(\dfrac{1}{2007}\)\(\dfrac{1}{2006}\)≠0)
=>x=2010
2 tháng 8 2021
Ta có:\(C=\left(\dfrac{1}{2}-1\right)\times\left(\dfrac{1}{3}-1\right)\times\left(\dfrac{1}{4}-1\right)\times...\times\left(\dfrac{1}{2006}-1\right)\times\left(\dfrac{1}{2007}-1\right)\)
\(=\dfrac{-1}{2}\times\dfrac{-2}{3}\times\dfrac{-3}{4}\times...\times\dfrac{-2005}{2006}\times\dfrac{-2006}{2007}\)
\(=\dfrac{1}{2007}\)
2 tháng 8 2021
Ta có: \(C=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2006}-1\right)\left(\dfrac{1}{2007}-1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot\dfrac{-3}{4}\cdot...\cdot\dfrac{-2005}{2006}\cdot\dfrac{-2006}{2007}\)
\(=\dfrac{1}{2007}\)
IM
1
6 tháng 7 2016
\(\frac{x+4}{2006}+\frac{x+3}{2007}=\frac{x+2}{2008}+\frac{x+1}{2009}\)
\(=>\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+1}{2009}+1\right)\)
\(=>\frac{x+2010}{2006}+\frac{x+2010}{2007}=\frac{x+2010}{2008}+\frac{x+2010}{2009}\)
\(=>\left(x+2010\right)\left(\frac{1}{2006}+\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)
\(=>x+2010=0\)
\(=>x=-2010\)
TN
2
21 tháng 2 2016
Số nguyên tố lớn hơn 3 chia 3 dư 1 hoặc dư 2
Vậy số nguyên tố lớn hơn 3 có dạng là:3k+1 hoặc 3k+2(k\(\in\)N*)
NK
11 tháng 1 2016
Theo bài ra, ta có:
x1 + x2 + x3 +...+ x2005 + x2006 + x2007 = 0 (1)
x1 + x2 = x3 + x4 =...= x2005 + x2006 = x2007 + x1 = 1 (2)
Từ (2) => x1 + x2 + x3 +...+ x2005 + x2006 + x2007 + x1 = 1 + 1 +...+ 1 (Đây là bước viết dãy đẳng thức thành tổng)
x1 + x2 + x3 +...+ x2005 + x2006 + x2007 + x1 = 1.1004
x1 + x2 + x3 +...+ x2005 + x2006 + x2007 + x1 = 1004
=> (x1 + x2 + x3 +...+ x2005 + x2006 + x2007 + x1) - (x1 + x2 + x3 +...+ x2005 + x2006 + x2007) = 1004 - 0
=> x1 = 1004 (Vì x1 là số bị thừa ra sau khi triệt tiêu)
Vậy...
9 tháng 1 2016
từ \(x_1\)+ \(x_2\) +........+ \(x_{2007}\)= 0
==>( x1 + x2) + ( x3+x4) +.......+ ( x2005 + x2006) + x2007= 0
==> 1+ 1 +.....+ 1 + x1007 = 0 ( 1003 số 1)
=> 1003 + x2007 = 0
=> x2007 = 0 - 1003
=> x2007 = -1003
vì x2007 + x1= 1 ==> -1003+ x1=1==> x1 = 1- 1003= -1002
Vậy x1 = -1002 ( tick nha)
GH
1
17 tháng 7 2022
\(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)
=>x-2010=0
hay x=2010
8 tháng 8 2016
1. 2006/987654321 + 2007/246813579 = 2007/246813579 + 2006/987654321
=>
2.
3 - (5.3/8 + X - 7 . 5/24) : 6 . 2/3 =2
3 - (15/8 + X - 35/24) : 4 = 2
3 - (15/8 + X - 35/24) = 2 . 4
3 - (15/8 + X - 35/24) = 8
15/8 + X - 35/24 = 3 - 8
15/8 + X - 35/24 = -5
15/8 + X = -5 + 35/24
15/8 + X = -85/24
X = -85/24 - 15/8
X = -65/12
\(x-\frac{x}{2}+\frac{x}{2}-\frac{x}{3}+...+\frac{x}{2006}-\frac{x}{2007}=\frac{2006}{2007}\)
\(x-\frac{x}{2007}=\frac{2006}{2007}\)
\(\frac{2007x-x}{2007}=\frac{2006}{2007}\)
\(\frac{2006x}{2007}=\frac{2006}{2007}\Rightarrow2006x=2006\)
=>x=1