K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
NT
1
1 tháng 4 2022
\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)
\(\Leftrightarrow\dfrac{x-1}{2013}-1+\dfrac{x-2}{2012}-1+\dfrac{x-3}{2011}-1=\dfrac{x-4}{2010}-1+\dfrac{x-5}{2009}-1+\dfrac{x-6}{2008}-1\)
=>x-2014=0
hay x=2014
PQ
2
ND
4 tháng 3 2018
\(\dfrac{x+1}{2012}+\dfrac{x+2}{2011}=\dfrac{x+3}{2010}+\dfrac{x+4}{2009}\)
\(\Leftrightarrow1+\dfrac{x+1}{2012}+1+\dfrac{x+2}{2011}=1+\dfrac{x+3}{2010}+1+\dfrac{x+4}{2009}\) \(\Leftrightarrow\dfrac{x+1+2012}{2012}+\dfrac{x+2+2011}{2011}=\dfrac{x+3+2010}{2010}+\dfrac{x+4+2009}{2009}\) \(\Leftrightarrow\dfrac{x+2013}{2012}+\dfrac{x+2013}{2011}-\dfrac{x+2013}{2010}-\dfrac{x+2013}{2009}=0\) \(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}\right)=0\)
Vì \(\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}\ne0\)
\(\Rightarrow x+2013=0\)
\(\Rightarrow x=-2013\)
Vậy........
4 tháng 3 2018
\(\dfrac{x+1}{2012}+\dfrac{x+2}{2011}=\dfrac{x+3}{2010}+\dfrac{x+4}{2009}\)
\(\Leftrightarrow\dfrac{x+1}{2012}+1+\dfrac{x+2}{2011}+1=\dfrac{x+3}{2010}+1+\dfrac{x+4}{2009}+1\)
\(\Leftrightarrow\dfrac{x+2013}{2012}+\dfrac{x+2013}{2011}-\dfrac{x+2013}{2010}-\dfrac{x+2013}{2009}=0\)
\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}\right)=0\)
\(\Leftrightarrow x=-2013\)(vì \(\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}\ne0\))
VT
0
LT
1
YN
2 tháng 3 2022
`Answer:`
\(\left(\frac{x+1}{2013}\right)+\left(\frac{x+2}{2012}\right)+\left(\frac{x+3}{2011}\right)=\left(\frac{x+4}{2010}\right)+\left(\frac{x+5}{2009}\right)+\left(\frac{x+6}{2008}\right)\)
\(\Leftrightarrow\frac{x+1}{2013}+1+\frac{x+2}{2012}+1+\frac{x+3}{2011}+1=\frac{x+4}{2010}+1+\frac{x+5}{2009}+1+\frac{x+6}{2008}+1\)
\(\Leftrightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}=\frac{x+2014}{2010}+\frac{x+2014}{2009}+\frac{x+2014}{2008}\)
\(\Leftrightarrow\frac{x+2014}{2013}+\frac{x+2014}{2012}+\frac{x+2014}{2011}-\frac{x+2014}{2010}-\frac{x+2014}{2009}-\frac{x+2014}{2008}=0\)
\(\Leftrightarrow\left(x+2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
\(\Rightarrow x+2014=0\)
\(\Leftrightarrow x=-2014\)
NT
3
G
11 tháng 3 2019
\(\frac{x+1}{2012}+\frac{x+2}{2011}=\frac{x+3}{2010}+\frac{x+4}{2009}\)
\(\Leftrightarrow\frac{x+1}{2012}+1+\frac{x+2}{2011}+1=\frac{x+3}{2010}+1+\frac{x+4}{2009}+1\)
\(\Leftrightarrow\frac{x+2013}{2012}+\frac{x+2013}{2011}=\frac{x+2013}{2010}+\frac{x+2013}{2009}\)
\(\Leftrightarrow\left(x+2013\right)\left(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}\right)=0\Leftrightarrow x=-2013\)
H
11 tháng 3 2019
\(\frac{x+1}{2012}+\frac{X+2}{2011}=\frac{X+3}{2010}+\frac{X+4}{2009}.\)
\(\Leftrightarrow\frac{X+1}{2012}+\frac{X+2}{2011}+2=\frac{X+3}{2010}+\frac{X+4}{2009}+2\)
\(\Leftrightarrow\frac{x+1}{2012}+1+\frac{x+2}{2011}+1=\frac{x+3}{2010}+1+\frac{x+4}{2009}+1\)
\(\Leftrightarrow\frac{x+2013}{2012}+\frac{x+2013}{2012}=\frac{x+2013}{2010}+\frac{x+2013}{2009}\)
\(\Leftrightarrow\left(x+2013\right).\left\{\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}\right\}=0\)
Mà \(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}>0\)
\(\Leftrightarrow x+2013=0\)
\(\Leftrightarrow x=-2013\)
KL ; PT có Nghiệm \(S=\left\{-2013\right\}\)
VT
2
TT
8 tháng 2 2020
\(\frac{x}{2008}+\frac{x+1}{2009}+...+\frac{x+4}{2012}=5\)
\(\Leftrightarrow\left(\frac{x}{2008}-1\right)+\left(\frac{x+1}{2009}-1\right)+...+\left(\frac{x+4}{2012}-1\right)=0\)
\(\Leftrightarrow\frac{x-2008}{2008}+\frac{x-2008}{2009}+...+\frac{x-2008}{2012}=0\)
\(\Leftrightarrow\left(x-2008\right)\left(\frac{1}{2008}+\frac{1}{2009}+..+\frac{1}{2012}\right)=0\)
Mà \(\left(\frac{1}{2008}+\frac{1}{2009}+..+\frac{1}{2012}\right)\ne0\)
Nên \(x-2008=0\)
\(\Leftrightarrow x=2008\)
Vậy : \(x=2008\)
CC
8 tháng 2 2020
\(\frac{x}{2008}+\frac{x+1}{2009}+\frac{x+2}{2010}+\frac{x+3}{2011}+\frac{x+4}{2012}=5\)
\(\Leftrightarrow\frac{x}{2008}+\frac{x+1}{2009}+\frac{x+2}{2010}+\frac{x+3}{2011}+\frac{x+4}{2012}-5=0\)
\(\Leftrightarrow\left(\frac{x}{2008}-1\right)+\left(\frac{x+1}{2009}-1\right)+\left(\frac{x+2}{2010}-1\right)+\left(\frac{x+3}{2011}-1\right)+\left(\frac{x+4}{2012}-1\right)=0\)
\(\Leftrightarrow\frac{x-2008}{2008}+\frac{x-2008}{2009}+\frac{x-2008}{2010}+\frac{x-2008}{2011}+\frac{x-2008}{2012}=0\)
\(\Leftrightarrow\left(x-2008\right)\left(\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)=0\)
Vì \(\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\ne0\)
\(\Rightarrow x-2008=0\)\(\Leftrightarrow x=2008\)
Vậy \(x=2008\)
NT
1
21 tháng 3 2016
=> 3x-(1/2013+2/2012+3/2011)=3x-(4/2010+5/2009+6/2008)=>6x=-4/2010-5/2009-6/2008+1/2013+2/2012+3/2011 =>x=... làm tiếp đi bạn
CN
1
Đề: Tính???
20122 - 20112 + 20102 - 20092 + ... + 22 - 12
= (2012 + 2011) (2012 - 2011) + (2010 + 2009) (2010 - 2009) + ... + (2 + 1) (2 - 1)
= 2012 + 2011 + 2010 + 2009 + ... + 2 + 1
Số số hạng là: (2012 - 1) : 1 + 1 = 2012 (số)
Tổng bằng: (1 + 2012) . 2012 : 2 = 2025078
Vậy 20122 - 20112 + 20102 - 20092 + ... + 22 - 12 = 2025078.