tìm x thuộc Z:
(2x−3)2015=(2x−3)2013(2x−3)2015=(2x−3)2013
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TK
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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
20 tháng 5 2022
a: Sửa đề: \(\dfrac{2x-1}{11}+\dfrac{2x-2}{12}+\dfrac{2x-3}{13}=\dfrac{2x+5}{5}+\dfrac{2x+7}{3}+\dfrac{2x+4}{6}\)
\(\Leftrightarrow\dfrac{2x-1}{11}+1+\dfrac{2x-2}{12}+1+\dfrac{2x-3}{13}+1=\dfrac{2x+5}{5}+1+\dfrac{2x+7}{3}+1+\dfrac{2x+4}{6}+1\)
=>2x+10=0
hay x=-5
b: \(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}+\dfrac{x-3}{2014}+\dfrac{x-4}{2013}+\dfrac{x-5}{2012}-5=0\)
\(\Leftrightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)+\left(\dfrac{x-3}{2014}-1\right)+\left(\dfrac{x-4}{2013}-1\right)+\left(\dfrac{x-5}{2012}-1\right)=0\)
=>x-2017=0
hay x=2017
20 tháng 5 2022
a: Sửa đề: \(\dfrac{2x-1}{11}+\dfrac{2x-2}{12}+\dfrac{2x-3}{13}=\dfrac{2x+5}{5}+\dfrac{2x+7}{3}+\dfrac{2x+4}{6}\)
\(\Leftrightarrow\dfrac{2x-1}{11}+1+\dfrac{2x-2}{12}+1+\dfrac{2x-3}{13}+1=\dfrac{2x+5}{5}+1+\dfrac{2x+7}{3}+1+\dfrac{2x+4}{6}+1\)
=>2x+10=0
hay x=-5
b: \(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}+\dfrac{x-3}{2014}+\dfrac{x-4}{2013}+\dfrac{x-5}{2012}-5=0\)
\(\Leftrightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)+\left(\dfrac{x-3}{2014}-1\right)+\left(\dfrac{x-4}{2013}-1\right)+\left(\dfrac{x-5}{2012}-1\right)=0\)
=>x-2017=0
hay x=2017
25 tháng 7 2018
i don't now
mong thông cảm !
...........................
\(\left(2x-3\right)^{2015}=\left(2x-3\right)^{2013}\)
\(\left(2x-3\right)^{2015}-\left(2x-3\right)^{2013}=0\)
\(\left(2x-3\right)^{2013}\text{[}\left(2x-3\right)^2-1\text{]}=0\)
=> \(\left(2x-3\right)^{2013}=0\) 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; \(\frac{3}{2}\)}