Tìm x : (x-5)^2016=(x-5)^2018
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DH
2
16 tháng 8 2023
(x – 5)2016 = (x – 5)2018
=> (x – 5)2018 – (x – 5)2016 = 0
=> (x – 5)2016.[(x – 5)2 – 1] = 0
=> x – 5 = 0 hoặc x – 5 = 1 hoặc x – 5 = -1
=> x = 5 hoặc x = 6 hoặc x = 4 (Thỏa mãn x ∈ N).
Vậy x ∈ {4; 5; 6}.
D
0
H
30 tháng 9 2023
Nghiệm?
\(\left(x-5\right)^{2016}+\left(x-5\right)^{2018}=0\\ \Rightarrow\left(x-5\right)^{2016}\left[1+\left(x-5\right)^2\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(x-5\right)^{2016}=0\\1+\left(x-5\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x\in\varnothing\end{matrix}\right.\)
NT
3
19 tháng 8 2018
Sửa lại đề:\(\frac{x+5}{2015}+\frac{x+4}{2016}=\frac{x+3}{2017}+\frac{x+2}{2018}\)
\(\frac{x+5}{2015}+1+\frac{x+4}{2016}+1=\frac{x+3}{2017}+1+\frac{x+2}{2018}+1\)
\(\frac{x+2020}{2015}+\frac{x+2020}{2016}=\frac{x+2020}{2017}+\frac{x+2020}{2018}\)
\(\frac{x+2020}{2015}+\frac{x+2020}{2016}-\frac{x+2020}{2017}-\frac{x+2020}{2018}=0\)
\(\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\right)=0\)
Do 1/2015+1/2016-1/2017-1/2018 khác 0
=>x+2020=0=>x=-2020
13 tháng 8 2019
a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\frac{x+2015}{5}+\frac{5}{5}+\frac{x+2016}{4}+\frac{4}{4}=\frac{x+2017}{3}+\frac{3}{3}+\frac{x+2018}{2}+\frac{2}{2}\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2002}{2}\)
\(\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right).\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\)
Vậy : \(x=-2020\)
Chúc bạn học tốt !!
13 tháng 8 2019
a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\\ \left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\\ \frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\\ \frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\\ \left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\\ \Rightarrow x+2020=0\\ \Rightarrow x=-2020\)
Vậy x = -2020
b) \(\frac{x+2015}{5}+\frac{x+2016}{6}=\frac{x+2017}{7}+\frac{x+2018}{8}\\ \left(\frac{x+2015}{5}-1\right)+\left(\frac{x+2016}{6}-1\right)=\left(\frac{x+2017}{7}-1\right)+\left(\frac{x+2018}{8}-1\right)\\ \frac{x+2010}{5}+\frac{x+2010}{6}=\frac{x+2010}{7}+\frac{x+2010}{8}\\ \frac{x+2010}{5}+\frac{x+2010}{6}-\frac{x+2010}{7}-\frac{x+2010}{8}=0\\ \left(x+2010\right)\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\right)=0\\ \Rightarrow x+2010=0\\ \Rightarrow x=-2010\)
Vậy x = -2010
Lời giải:
$(x-5)^{2018}=(x-5)^{2016}$
$(x-5)^{2018}-(x-5)^{2016}=0$
$(x-5)^{2016}[(x-5)^2-1)=0$
$\Rightarrow (x-5)^{2016}=0$ hoặc $(x-5)^2=1$
$\Rightarrow x-5=0$ hoặc $x-5=1$ hoặc $x-5=-1$
$\Rightarrow x=5$ hoặc $x=6$ hoặc $x=4$
\(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\)
\(\Leftrightarrow\left(x-5\right)^{2016}-\left(x-5\right)^{2018}=0\)
\(\Leftrightarrow\left(x-5\right)^{2016}\left[1-\left(x-5\right)^2\right]=0\)
\(\Leftrightarrow\left(x-5\right)^{2016}\left(1+x-5\right)\left(1-x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-4=0\\6-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)