Tìm x biết: \(\dfrac{x-1016}{1001}+\dfrac{x-13}{1002}+\dfrac{x+992}{1003}=\dfrac{x+995}{1004}+\dfrac{x-7}{1005}+1\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
ND
25 tháng 3 2018
\(\dfrac{x-1001}{1006}+\dfrac{x-1003}{1004}+\dfrac{x-1005}{1002}+\dfrac{x-1007}{1000}=4\)
\(\Leftrightarrow\dfrac{x-1001}{1006}-1+\dfrac{x-1003}{1004}-1+\dfrac{x-1005}{1002}-1+\dfrac{x-1007}{1000}-1=0\)
\(\Leftrightarrow\dfrac{x-2007}{1006}+\dfrac{x-2007}{1004}+\dfrac{x-2007}{1002}+\dfrac{x-2007}{1000}=0\)
\(\Leftrightarrow\left(x-2007\right)\left(\dfrac{1}{1006}+\dfrac{1}{1004}+\dfrac{1}{1002}+\dfrac{1}{1000}=0\right)\)
\(\Leftrightarrow x-2007=0\)
\(\Leftrightarrow x=2007\)
PM
1
NM
Nguyễn Minh Quang
Giáo viên
18 tháng 3 2022
ta có :
\(\frac{x-1009}{1001}-1+\frac{x-4}{1003}-2+\frac{x+2010}{1005}-4=0\)
hay \(\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\Leftrightarrow x-2010=0\)
hay x =2010
Vậy phương trình có nghiệm x = 2010
HT
2
QD
19 tháng 5 2018
\(\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}=7\)
\(\Leftrightarrow\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}-7=0\)
\(\Leftrightarrow\left(\dfrac{x-1009}{1001}-1\right)+\left(\dfrac{x-4}{1003}-2\right)+\left(\dfrac{x+2010}{1005}-4\right)=0\)
\(\Leftrightarrow\dfrac{x-2010}{1001}+\dfrac{x-2010}{1003}+\dfrac{x-2010}{1005}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{1001}+\dfrac{1}{1003}+\dfrac{1}{1005}\right)=0\)
\(\Leftrightarrow x-2010=0\)
\(\Rightarrow x=2010\)
Vậy....
21 tháng 7 2018
\(\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}=7\)
\(\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}-7=0\)
\(\left(\dfrac{x-1009}{1001}-1\right)+\left(\dfrac{x-4}{1003}-2\right)+\left(\dfrac{x+2010}{1005}-4\right)=0\)
\(\dfrac{x-2010}{1001}+\dfrac{x-2010}{1003}+\dfrac{x-2010}{1005}=0\)
\(\left(x-2010\right)\left(\dfrac{1}{1001}+\dfrac{1}{1003}+\dfrac{1}{1005}\right)=0\)
\(x-2010=0\)
\(x=2010\)
Vậy x = 2010
3 tháng 3 2017
\(\frac{x+1006}{1007}+\frac{x+1005}{1008}=\frac{x+1004}{1009}+\frac{x+1003}{1010}\)
\(\Rightarrow\left(\frac{x+1006}{1007}+1\right)+\left(\frac{x+1005}{1008}+1\right)=\left(\frac{x+1004}{1009}+1\right)+\left(\frac{x+1003}{1010}+1\right)\)
\(\Rightarrow\frac{x+2013}{1007}+\frac{x+2013}{1008}=\frac{x+2013}{1009}+\frac{x+2013}{1010}\)
\(\Rightarrow\frac{x+2013}{1007}+\frac{x+2013}{1008}-\frac{x+2013}{1009}-\frac{x+2013}{1010}=0\)
\(\Rightarrow\left(x+2013\right)\left(\frac{1}{1007}+\frac{1}{1008}-\frac{1}{1009}-\frac{1}{1010}\right)=0\)
Mà \(\frac{1}{1007}+\frac{1}{1008}-\frac{1}{1009}-\frac{1}{1010}\ne0\)
\(\Rightarrow x+2013=0\)
\(\Rightarrow x=-2013\)
Vậy x = -2013
CV
8 tháng 4 2018
ta thấy : \(\dfrac{-1003}{-2002}\) = \(\dfrac{1003}{2002}\)
\(\dfrac{1004}{-2003}\) = \(\dfrac{-1004}{2003}\)
Sắp xếp : \(\dfrac{1004}{-2003}\) <\(\dfrac{-1003}{2003}\) <\(\dfrac{-1002}{2003}\) <\(\dfrac{1001}{2002}\) <\(\dfrac{-1003}{-2002}\)
HT
1
6 tháng 12 2017
\(\dfrac{x-1001}{1006}+\dfrac{x-1003}{1004}+\dfrac{x-1005}{1002}+\dfrac{x-1007}{1000}=4\)
\(\Rightarrow\dfrac{x-1001}{1006}-1+\dfrac{x-1003}{1004}-1+\dfrac{x-1005}{1002}-1+\dfrac{x-1007}{1000}-1=0\)
\(\Rightarrow\dfrac{x-2007}{1006}+\dfrac{x-2007}{1004}+\dfrac{x-2007}{1002}+\dfrac{x-2007}{1000}=0\)
\(\Rightarrow\left(x-2007\right)\left(\dfrac{1}{1006}+\dfrac{1}{1004}+\dfrac{1}{1002}+\dfrac{1}{1000}\right)=0\)
Dễ thấy: \(\dfrac{1}{1000}+\dfrac{1}{1004}+\dfrac{1}{1002}+\dfrac{1}{1000}>0\Leftrightarrow x-2007=0\Leftrightarrow x=2007\)
TT
26 tháng 5 2021
a) \(P=\dfrac{x\left(x-4\right)+4}{x^2-4}\)
\(P=\dfrac{x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\)
\(P=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\)
\(P=\dfrac{x-2}{x+2}\)
b) \(P=\dfrac{x-2}{x+2}\)=\(\dfrac{1001}{1003}\)
\(=>1003x-2006=1001x+2002\)
\(2x=4008=>x=2004\)
27 tháng 5 2021
a) ĐK: \(x\ne2;x\ne-2\)
\(P=\dfrac{x\left(x-4\right)+4}{x^2-4}
=\dfrac{x^2-4x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x-2}{x+2}\)
b) \(P=\dfrac{1001}{1003}\Leftrightarrow\dfrac{x-2}{x+2}=\dfrac{1001}{1003}\Rightarrow1003x-2006=1001x+2002\)
\(\Leftrightarrow2x=4008\Leftrightarrow x=2004\) (tmđk)
LT
1
H
5 tháng 4 2023
\(đk:x\ne0;x\ne-1\\ \dfrac{3}{x}+\dfrac{x}{x+1}+\dfrac{x-3}{x}=\dfrac{13}{7}\\ \Leftrightarrow\dfrac{7.3.\left(x+1\right)}{7x\left(x+1\right)}+\dfrac{x.x.7}{7x\left(x+1\right)}+\dfrac{\left(x-3\right)\left(x+1\right).7}{7x\left(x+1\right)}=\dfrac{13.x.\left(x+1\right)}{7x\left(x+1\right)}\\ \Leftrightarrow\dfrac{21x+21+7x^2+7x^2+7x-21x-21}{7x\left(x+1\right)}=\dfrac{13x^2+13x}{7x\left(x+1\right)}\\ \Leftrightarrow14x^2+7x=13x^2+13x\\ \Leftrightarrow14x^2-13x^2=13x-7x\\ \Leftrightarrow x^2=6x\\ \Leftrightarrow x^2-6x=0\\ \Leftrightarrow x\left(x-6\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(kot/m\right)\\x=6\left(t/m\right)\end{matrix}\right.\Rightarrow x=6\)
\(\dfrac{x-1016}{1001}+\dfrac{x-13}{1002}+\dfrac{x+992}{1003}=\dfrac{x+995}{1004}+\dfrac{x-7}{1005}+1\)
<=>\(\dfrac{x-1016}{1001}-1+\dfrac{x-13}{1002}-2+\dfrac{x+992}{1003}-3=\dfrac{x+995}{1004}-3+\dfrac{x-7}{1005}-2\)
<=>\(\dfrac{x-2017}{1001}+\dfrac{x-2017}{1002}+\dfrac{x-2017}{1003}=\dfrac{x-2017}{1004}+\dfrac{x-2017}{1005}\)
<=>\(\left(x-2017\right)\left(\dfrac{1}{1001}+\dfrac{1}{1002}+\dfrac{1}{1003}-\dfrac{1}{1004}-\dfrac{1}{1005}\right)=0\)
vì 1/1001+1/1002+1/1003-1/1004-1/1005 khác 0 nên x-2017=0<=>x=2017
vậy..........