Giải phương trình x + 1 2012 - x + 2 2011 = x + 3 2010 + 1
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CM
Cao Minh Tâm
11 tháng 4 2019
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NM
2
7 tháng 5 2018
Ta có :
\(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+...+\frac{x-2012}{1}=2012\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+\left(\frac{x-3}{2010}-1\right)+...+\left(\frac{x-2012}{1}-1\right)=2012\)
\(\Leftrightarrow\)\(\frac{x-1-2012}{2012}+\frac{x-2-2011}{2011}+\frac{x-3-2010}{2010}+...+\frac{x-2012-1}{1}=0\)
\(\Leftrightarrow\)\(\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\)\(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{1}\right)=0\)
Vì \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{1}\ne0\)
Nên \(x-2013=0\)
\(\Leftrightarrow\)\(x=2013\)
Vậy \(x=2013\)
Chúc bạn học tốt ~
7 tháng 5 2018
\(\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+...+\frac{x-2012}{1}-1+2012=2012\)
\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{1}\right)=0\)
\(\Leftrightarrow x=2013\)
7 tháng 1 2018
Phương trình đã cho tương đương với :
\(\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+\frac{x-3}{2010}-1+...+\frac{x-2012}{1}-1+2012=2012\)
\(\Leftrightarrow\)\(\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\)\(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{1}\right)=0\)
Tìm x theo như toán lớp 6 nha
\(x-2013=0\)
\(\Leftrightarrow\)\(x=2013\)
VT
7 tháng 1 2018
ta có pt
<=>\(\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+...+\frac{x-2012}{1}-1=0\)
<=>\(\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)
<=>\(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{1}\right)=0\Leftrightarrow x-2013=0\Leftrightarrow x=2013\)
^_^
VT
0
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 \(x=2008\)
CN
12 tháng 2 2019
\(\Leftrightarrow\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+...+\left(\frac{x-2012}{1}-1\right)=0\)
\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+....+1\right)=0\)
\(\Leftrightarrow x-2013=0\)(because 1/2012 +1/2011+...+1 luôn lớn hơn 0
\(\Leftrightarrow x=2013\)
Vậy ........
TA
28 tháng 3 2021
`(x-1)/2013+(x-2)/2012+(x-3)/2011=(x-4)/2010+(x-5)/2009 +(x-6)/2008`
`<=> ((x-1)/2013-1)+((x-2)/2012-1)+((x-3)/2011-1)=( (x-4)/2010-1)+((x-5)/2009-1)+((x-6)/2008-1)`
`<=> (x-2014)/2013 +(x-2014)/2012+(x-2014)/2011=(x-2014)/2010+(x-2014)/2009+(x-2014)/2008`
`<=> x-2014=0` (Vì `1/2013+1/2012+1/2011-1/2010-1/2009-1/2008 \ne 0`)
`<=>x=2014`
Vậy `S={2014}`.
17 tháng 3 2023
\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2013}-1\right)+\left(\dfrac{x-2}{2012}-1\right)+\left(\dfrac{x-3}{2011}-1\right)=\left(\dfrac{x-4}{2010}-1\right)+\left(\dfrac{x-5}{2009}-1\right)+\left(\dfrac{x-6}{2008}-1\right)\)
\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}=\dfrac{x-2014}{2010}+\dfrac{x-2014}{2009}+\dfrac{x-2014}{2008}\)
\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}-\dfrac{x-2014}{2010}-\dfrac{x-2014}{2009}-\dfrac{x-2014}{2008}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)
\(\Leftrightarrow\left(x-2014\right).A=0\)
\(\text{Vì A }\ne0\)
\(\Rightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{2014\right\}\)
22 tháng 4 2020
Bài 1 :
Ta có :
\(\frac{x+2011}{2013}+\frac{x+2012}{2012}=\frac{x+2010}{2014}+\frac{x+2013}{2011}\)
\(\Rightarrow\left(\frac{x+2011}{2013}+1\right)+\left(\frac{x+2012}{2012}+1\right)=\left(\frac{x+2010}{2014}+1\right)\)
\(+\left(\frac{x+2013}{2011}+1\right)\)
\(\Rightarrow\frac{x+4024}{2013}+\frac{x+4024}{2012}=\frac{x+4024}{2014}+\frac{x+4024}{2011}\)
\(\Rightarrow\frac{x+4024}{2013}+\frac{x+4024}{2012}-\frac{x+4024}{2014}-\frac{x+4024}{2011}=0\)
\(\Rightarrow\left(x+4024\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2014}-\frac{1}{2011}\right)=0\)
\(\Rightarrow x+4024=0\)
\(\Rightarrow x=-4024\)
22 tháng 4 2020
Bài 2 :
Đặt \(x^2+2x+1=a\Rightarrow a=\left(x+1\right)^2\ge0\)
=> Phương trình trở thành
\(\frac{a}{a+1}+\frac{a+1}{a+2}=\frac{7}{6}\)
\(\Rightarrow\frac{a}{a+1}.6\left(a+1\right)\left(a+2\right)+\frac{a+1}{a+2}.6\left(a+1\right)\left(a+2\right)=\frac{7}{6}.6\left(a+1\right)\left(a+2\right)\)
\(\Rightarrow6a\left(a+2\right)+6\left(a+1\right)^2=7\left(a+1\right)\left(a+2\right)\)
\(\Rightarrow12a^2+24a+6=7a^2+21a+14\)
\(\Rightarrow5a^2+3a-8=0\)
\(\Rightarrow\left(a-1\right)\left(5a+8\right)=0\)
Vì \(a\ge0\Rightarrow a=1\)
\(\Rightarrow x^2+2x+1=1\)
\(x^2+2x=0\)
\(\Rightarrow x\left(x+2\right)=0\)
\(\Rightarrow x\in\left\{-2,0\right\}\)
