Giải phương trình x-3/2013 + x-2/2014 = x-2014/2 + x-2013/3
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
19 tháng 1 2018
\(\dfrac{x-3}{2013}+\dfrac{x-2}{2014}=\dfrac{x-2014}{2}+\dfrac{x-2013}{3}\)
\(\Leftrightarrow\left(\dfrac{x-3}{2013}-1\right)+\left(\dfrac{x-2}{2014}-1\right)=\left(\dfrac{x-2014}{2}-1\right)+\left(\dfrac{x-2013}{3}-1\right)\)\(\Leftrightarrow\dfrac{x-2016}{2013}+\dfrac{x-2016}{2014}=\dfrac{x-2016}{2}+\dfrac{x-2016}{3}\)
\(\Leftrightarrow\left(x-2016\right)\left(\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
Hiển nhiên \(\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2}-\dfrac{1}{3}< 0.\)
Vậy \(S=\left\{2016\right\}\)
TA
19 tháng 1 2018
\(\dfrac{x-3}{2013}-1+\dfrac{x-2}{2014}-1=\dfrac{x-2014}{2}-1+\dfrac{x-2013}{3}-1\)
=> \(\dfrac{x-3-2013}{2013}+\dfrac{x-2-2014}{2014}=\dfrac{x-2014-2}{2}+\dfrac{x-2013-3}{3}\)
=> \(\dfrac{x-2016}{2013}+\dfrac{x-2016}{2014}-\dfrac{x-2016}{2}-\dfrac{x-2016}{3}=0\)
=> (x-2016)\(\left(\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
=> x-2016=0
=> x=2016
1 tháng 1
1)
\(\dfrac{x-1}{2014}+\dfrac{x-2}{2013}+\dfrac{x-3}{2012}+...+\dfrac{x-2014}{1}=2014\)
\(\Leftrightarrow\left(\dfrac{x-1}{2014}-1\right)+\left(\dfrac{x-2}{2013}-1\right)+...+\left(\dfrac{x-2014}{1}-1\right)=0\)
\(\Leftrightarrow\dfrac{x-2015}{2014}+\dfrac{x-2015}{2013}+...+\dfrac{x-2015}{1}=0\)
\(\Leftrightarrow\left(x-2025\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1}\right)=0\)
\(\Leftrightarrow x=2015\)
Vậy \(S=\left\{2015\right\}\)
14 tháng 12 2023
Điều kiện: \(x\ge2012;y\ge2013;z\ge2014\)
Áp dụng bất đẳng thức Cauchy, ta có:
\(\left\{{}\begin{matrix}\dfrac{\sqrt{x-2012}-1}{x-2012}=\dfrac{\sqrt{4\left(x-2012\right)}-2}{2\left(x-2012\right)}\le\dfrac{\dfrac{4+x-2012}{2}-2}{2\left(x-2012\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{y-2013}-1}{y-2013}=\dfrac{\sqrt{4\left(y-2013\right)}-2}{2\left(y-2013\right)}\le\dfrac{\dfrac{4+y-2013}{2}-2}{2\left(y-2013\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{z-2014}-1}{z-2014}=\dfrac{\sqrt{4\left(z-2014\right)}-2}{2\left(z-2014\right)}\le\dfrac{\dfrac{4+z-2014}{2}-2}{2\left(z-2014\right)}=\dfrac{1}{4}\end{matrix}\right.\)
Cộng vế theo vế, ta được:
\(\dfrac{\sqrt{x-2012}-1}{x-2012}+\dfrac{\sqrt{y-2013}-1}{y-2013}+\dfrac{\sqrt{z-2014}-1}{z-2014}\le\dfrac{3}{4}\)
Đẳng thức xảy ra khi \(x=2016;y=2017;z=2018\)
Vậy....
KE
2
8 tháng 3 2018
pt <=> (x/2012 - 1) + (x+1/2013 - 1) + (x+2/2014 - 1) + (x+3/2015 - 1) + (x+4/2016 - 1) = 0
<=> x-2012/2012 + x-2012/2013 + x-2012/2014 + x-2012/2015 + x-2012/2016 = 0
<=> (x-2012).(1/2012+1/2013+1/2014+1/2015+1/2016) = 0
<=> x-2012 = 0 ( vì 1/2012+1/2013+1/2014+1/2015+1/2016 > 0 )
<=> x=2012
Vậy x=2012
Tk mk nha
8 tháng 3 2018
Ta có :
\(\frac{x}{2012}+\frac{x+1}{2013}+\frac{x+2}{2014}+\frac{x+3}{2015}+\frac{x+4}{2016}=5\)
\(\Leftrightarrow\)\(\left(\frac{x}{2012}-1\right)+\left(\frac{x+1}{2013}-1\right)+\left(\frac{x+2}{2014}-1\right)+\left(\frac{x+3}{2015}-1\right)+\left(\frac{x+4}{2016}-1\right)=5-5\)
\(\Leftrightarrow\)\(\frac{x-2012}{2012}+\frac{x-2012}{2013}+\frac{x-2012}{2014}+\frac{x-2012}{2015}+\frac{x-2012}{2016}=0\)
\(\Leftrightarrow\)\(\left(x-2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)=0\)
Vì \(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\ne0\)
\(\Rightarrow\)\(x-2012=0\)
\(\Rightarrow\)\(x=2012\)
Vậy \(x=2012\)
Chúc bạn học tốt ~
23 tháng 10 2016
Đặt \(a=2x^2+x-2014\) , \(b=x^2-5x-2013\)
thì \(a^2+4b^2=4ab\Leftrightarrow a^2-4ab+4b^2=0\Leftrightarrow\left(a-2b\right)^2=0\)
Thay vào được \(\left[\left(2x^2+x-2014\right)-2\left(x^2-5x-2013\right)\right]^2=0\)
\(\Leftrightarrow11x+2012=0\Leftrightarrow x=-\frac{2012}{11}\)
NT
0
D
2
20 tháng 11 2017
Cộng 2 vế với 2 ta có :
5-x^2/2012 + 1 = (4-x^2/2013+1) - (x^2-3/2014-1)
<=> 2017-x^2/2012 = 2017-x^2/2013 - x^2-2017/2014 = 2017-x^2/2013+ 2017-x^2/2014
<=> 2017-x^2/2013 + 2017-x^2/2014 - 2017-x^2/2012 = 0
<=> (2017-x^2).(1/2013+1/2014-1/2012) = 0
<=> 2017-x^2 = 0 ( vì 1/2013+1/2014-1/2012 khác 0 )
<=> x = \(\sqrt{2017}\)
k mk nha
4 tháng 3 2019
\(\Leftrightarrow\frac{5-x^2}{2012}+1=\frac{4-x^2}{2013}+1+\frac{3-x^2}{2014}+1\)
\(\Leftrightarrow\frac{2017-x^2}{2012}-\frac{2017-x^2}{2013}-\frac{2017-x^2}{2014}=0\)
\(\Leftrightarrow\left(2017-x^2\right)\left(\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
\(\Leftrightarrow2017-x^2=0\)
\(\Leftrightarrow x^2=2017\)
\(\Leftrightarrow x=\sqrt{2017}\)
V...\(S=\left\{\sqrt{2017}\right\}\)
11 tháng 1 2017
Theo bài ra , ta có :
\(\frac{x+2}{2014}+\frac{x+1}{2015}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\Leftrightarrow\left(\frac{x+2}{2014}+1\right)+\left(\frac{x+1}{2015}+1\right)=\left(\frac{x+3}{2013}+1\right)+\left(\frac{x+4}{2012}+1\right)\)
\(\Leftrightarrow\left(\frac{x+2+2014}{2014}\right)+\left(\frac{x+1+2015}{2015}\right)=\left(\frac{x+3+2013}{2013}\right)+\left(\frac{x+4+2012}{2012}\right)\)
\(\Leftrightarrow\frac{x+2016}{2014}+\frac{x+2016}{2015}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)
\(\Leftrightarrow\frac{x+2016}{2014}+\frac{x+2016}{2015}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\)
\(\Leftrightarrow\left(x+2016\right)\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
Vì \(\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)>0\)
\(\Leftrightarrow x+2016=0\)
\(\Leftrightarrow x=-2016\)
Vậy \(x=-2016\)
Tập nghiệm của phương trình là \(S=\left\{-2016\right\}\)
Chúc bạn học tốt =))
11 tháng 1 2017
\(\frac{x+2}{2014}+\frac{x+1}{2015}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(\frac{x+2}{2014}+1+\frac{x+1}{2015}+1=\frac{x+3}{2013}+1+\frac{x+4}{2012}+1\)
\(\frac{x+2+2014}{2014}+\frac{x+1+2015}{2015}=\frac{x+3+2013}{2013}+\frac{x+4+2012}{2012}\)
\(\frac{x+2016}{2014}+\frac{x+2016}{2015}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)
\(\frac{x+2016}{2014}+\frac{x+2016}{2015}-\frac{x+2016}{2013}-\frac{x+2016}{2012}=0\)
\(\left(x+2016\right).\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)
MÀ \(\left(\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2012}\right)\ne0\)
\(\Rightarrow x+2016=0\)
\(\Rightarrow x=-2016\)
\(\dfrac{x-3}{2013}+\dfrac{x-2}{2014}=\dfrac{x-2014}{2}+\dfrac{x-2013}{3}\)
\(\Leftrightarrow\dfrac{x-3}{2013}-1+\dfrac{x-2}{2014}-1=\dfrac{x-2014}{2}-1+\dfrac{x-2013}{3}-1\)
\(\Leftrightarrow\dfrac{x-3-2013}{2013}+\dfrac{x-2-2014}{2014}=\dfrac{x-2014-2}{2}+\dfrac{x-2013-3}{3}\)
\(\Leftrightarrow\dfrac{x-2016}{2013}+\dfrac{x-2016}{2014}=\dfrac{x-2016}{2}+\dfrac{x-2016}{3}\)
\(\Leftrightarrow\dfrac{x-2016}{2013}+\dfrac{x-2016}{2014}-\dfrac{x-2016}{2}-\dfrac{x-2016}{3}=0\)
\(\Leftrightarrow\left(x-2016\right)\left(\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
Vì \(\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2}-\dfrac{1}{3}\ne0\)
\(\Rightarrow x-2016=0\)
\(\Leftrightarrow x=2016\)( thỏa mãn )
Vậy x = 2016
\(\dfrac{x-3}{2013}+\dfrac{x-2}{2014}=\dfrac{x-2014}{2}+\dfrac{x-2013}{3}\)
\(\Leftrightarrow\dfrac{x-3}{2013}-1+\dfrac{x-2}{2014}-1=\dfrac{x-2014}{2}-1+\dfrac{x-2013}{3}-1\)
\(\Leftrightarrow\dfrac{x-2016}{2013}+\dfrac{x-2016}{2014}=\dfrac{x-2016}{2}+\dfrac{x-2016}{3}\)
\(\Leftrightarrow\left(x-2016\right)\left(\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow x-2016=0\) (do \(\dfrac{1}{2013}+\dfrac{1}{2014}-\dfrac{1}{2}-\dfrac{1}{3}\ne0\))
\(\Rightarrow x=2016\)