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TK
3
13 tháng 7 2017
\(\left(2x-3\right)^{2015}=\left(2x-3\right)^{2013}\)
\(\Rightarrow\left(2x-3\right)^{2015}-\left(2x-3\right)^{2013}=0\)
\(\Rightarrow\left(2x-3\right)^{2013}\left[\left(2x-3\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x-3\right)^{2013}=0\\\left(2x-3\right)^2-1=0\end{matrix}\right.\)
+) \(\left(2x-3\right)^{2013}=0\Rightarrow x=\dfrac{3}{2}\)
+) \(\left(2x-3\right)^2-1=0\Rightarrow\left[{}\begin{matrix}2x-3=1\\2x-3=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
Vậy \(x=\dfrac{3}{2}\) hoặc x = 2 hoặc x = 1
DH
15 tháng 8 2017
\(\dfrac{1-2x}{2017}+\dfrac{2-2x}{2016}=\dfrac{3-2x}{2015}+\dfrac{4-2x}{2014}\)
\(\Rightarrow\left(\dfrac{1-2x}{2017}+1\right)+\left(\dfrac{2-2x}{2016}+1\right)=\left(\dfrac{3-2x}{2015}+1\right)+\left(\dfrac{4-2x}{2014}+1\right)\)
\(\Rightarrow\dfrac{2018-2x}{2017}+\dfrac{2018-2x}{2016}-\dfrac{2018-2x}{2015}-\dfrac{2018-2x}{2014}=0\)
\(\Rightarrow\left(2018-2x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Vì \(2017>2016>2015>2014\) nên
\(\dfrac{1}{2017}< \dfrac{1}{2016}< \dfrac{1}{2015}< \dfrac{1}{2014}\)
\(\Rightarrow\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}< 0\)
\(\Rightarrow2018-2x=0\Rightarrow x=1009\)
Vậy...........
Chúc bạn học tốt!!!
HD
15 tháng 8 2017
\(\dfrac{1-2x}{2017}+\dfrac{2-2x}{2016}=\dfrac{3-2x}{2015}+\dfrac{4-2x}{2014}\)
\(\Rightarrow\left(\dfrac{1-2x}{2017}+1\right)+\left(\dfrac{2-2x}{2016}+1\right)=\left(\dfrac{3-2x}{2015}+1\right)+\left(\dfrac{4-2x}{2014}+1\right)\)
\(\Rightarrow\dfrac{2018-2x}{2017}+\dfrac{2018-2x}{2016}-\dfrac{2018-2x}{2015}-\dfrac{2018-2x}{2014}=0\)
\(\Rightarrow\left(20418-2x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
\(Ta\) \(có\)\(:\) \(2017>2016>2015>2014\)
\(\Rightarrow\dfrac{1}{2017}< \dfrac{1}{2016}< \dfrac{1}{2015}< \dfrac{1}{2014}\)
\(\Rightarrow\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}< 0\)
\(\Rightarrow2018-2x=0\)
\(\Rightarrow2x=2018-0\)
\(\Rightarrow2x=2018\)
\(\Rightarrow x=2018:2\)
\(\Rightarrow x=1009\)
PC
0
HT
2
20 tháng 1 2019
Ta có : ( 2x - 6 ) 2013 = ( 2x - 6 ) 3
( 2x - 6 ) 2013 - ( 2x - 6 ) 3 = 0
( 2x - 6 ) 3 . [ ( 2x - 6 ) 2010 - 1 ] = 0
( 2x - 6 ) 3 = 0 ; ( 2x - 6 ) 2010 = 1
2x - 6 = 0 ; 2x - 6 = -1 ; 2x - 6 = 1
2x = 6 ; 2x = 5 ; 2x = 7
x = 3 ; x = 2,5 ; x = 3,5
Vậy x = 2,5 ; 3 ; 3,5
20 tháng 1 2019
\(\left(2x-6\right)^{2013}=\left(2x-6\right)^3\)
\(\Rightarrow\left(2x-6\right)^{2013}-\left(2x-6\right)^3=0\)
\(\Rightarrow\left(2x-6\right)^3\left[\left(2x-6\right)^{2010}-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(2x-6\right)^3=0\\\left(2x-6\right)^{2010}-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-6=0\\\left(2x-6\right)^{2010}=1\end{cases}}\)
<=> hoặc 2x-6=0 hoặc 2x-6=1 hoặc 2x-6=-1
<=> hoặc x=3 hoặc x=7/2 hoặc x=5/2
\(\left(2x-3\right)^{2015}=\left(2x-3\right)^{2013}\)
\(\left(2x-3\right)^{2015}-\left(2x-3\right)^{2013}=0\)
\(\left(2x-3\right)^{2013}\text{[}\left(2x-3\right)^2-1\text{]}=0\)
=> \(\left(2x-3\right)^{2013}=0\) hoặc \(\left(2x-3\right)^2-1=0\)
+) => 2x-3=0=>2x=3=>x=\(\frac{3}{2}\)
+) => \(\left(2x-3\right)^2\)=1
_ 2x-3=1 _2x-3=-1
=>2x=4 =>2x=2
=>x=2 =>x=1
Vậy x={2; 1; \(\frac{3}{2}\)}