\(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\)
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
14 tháng 12 2016
\(\left(x-\frac{1}{5}\right)^{2014}+\left(y+0,4\right)^{2016}+\left(z-3\right)^{2018}=0\)
Ta thấy: \(\begin{cases}\left(x-\frac{1}{5}\right)^{2014}\ge0\\\left(y+0,4\right)^{2016}\ge0\\\left(z-3\right)^{2018}\ge0\end{cases}\)
\(\Rightarrow\left(x-\frac{1}{5}\right)^{2014}+\left(y+0,4\right)^{2016}+\left(z-3\right)^{2018}\ge0\)
\(\Rightarrow\begin{cases}\left(x-\frac{1}{5}\right)^{2014}=0\\\left(y+0,4\right)^{2016}=0\\\left(z-3\right)^{2018}=0\end{cases}\)\(\Rightarrow\begin{cases}x-\frac{1}{5}=0\\y+0,4=0\\z-3=0\end{cases}\)\(\Rightarrow\begin{cases}x=\frac{1}{5}\\y=-0,4\\z=3\end{cases}\)
DT
1
21 tháng 4 2017
Dễ thấy: \(\left\{{}\begin{matrix}\left|x+2016\right|\ge0\\\left|x+2017\right|\ge0\\\left|x+2018\right|\ge0\end{matrix}\right.\)\(\forall x\)
\(\Rightarrow\left|x+2016\right|+\left|x+2017\right|+\left|x+2018\right|\ge0\forall x\)
\(\Rightarrow VT\ge0\Rightarrow VP\ge0\Rightarrow10x\ge0\Rightarrow x\ge10\)
\(pt\Leftrightarrow\left(x+2016\right)+\left(x+2017\right)+\left(x+2018\right)=10x\)
\(\Leftrightarrow3x+6051=10x\)
\(\Leftrightarrow6051=7x\Rightarrow x=\dfrac{6051}{7}\)
DQ
1
VM
1 tháng 10 2019
\(\left(2x-5\right)^{2016}+\left(3y+4\right)^{2018}\le0\)
Ta có:
\(\left\{{}\begin{matrix}\left(2x-5\right)^{2016}\ge0\\\left(3y+4\right)^{2018}\ge0\end{matrix}\right.\forall x.\)
\(\Rightarrow\left(2x-5\right)^{2016}+\left(3y+4\right)^{2018}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-5\right)^{2016}=0\\\left(3y+4\right)^{2016}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\3y+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=0+5=5\\3y=0-4=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5:2\\y=\left(-4\right):3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\y=-\frac{4}{3}\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{\frac{5}{2};-\frac{4}{3}\right\}.\)
Chúc bạn học tốt!
ST
1
30 tháng 3 2018
Với mọi x ta có :
\(\left|x+2018\right|=\left|-x-2018\right|\)
\(\Leftrightarrow\left|x+2016\right|+\left|x+2018\right|=\left|x+2016\right|+\left|-x-2018\right|\)
\(\Leftrightarrow\left|x+2016\right|+\left|-x-2018\right|\ge\left|\left(x+2016\right)+\left(-x-2018\right)\right|\)
\(\Leftrightarrow\left|x+2016\right|+\left|-x-2018\right|\ge\left|-2\right|\)
\(\Leftrightarrow\left|x+2016\right|+\left|-x-2018\right|\ge2\)
Mà \(\left|x+2017\right|\ge0\)
\(\Leftrightarrow\left|x+2016\right|+\left|-x-2018\right|+\left|x+2017\right|\ge2\)
Dấu "=" xảy ra khi :
\(\left\{{}\begin{matrix}\left(x+2016\right)\left(-x-2018\right)\ge0\left(1\right)\\\left|x+2017\right|=0\left(2\right)\end{matrix}\right.\)
Từ \(\left(1\right)\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+2016\ge0\\-x-2018\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x+2016\le0\\-x-2018\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-2016\\-2018\ge x\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-2016\\-2018\le x\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}-2016\ge x\ge-2018\\x\in\varnothing\end{matrix}\right.\)
\(\Leftrightarrow-2016\ge x\ge-2018\left(I\right)\)
Từ \(\left(2\right)\Leftrightarrow x+2017=0\)
\(\Leftrightarrow x=-2017\left(II\right)\)
Từ \(\left(I\right)+\left(II\right)\Leftrightarrow GTNN\) của \(\left|x+2016\right|+\left|x+2017\right|+\left|x+2017\right|=2\Leftrightarrow x=-2017\)
L1
0
BT
1
\(\left(x-5\right)^{2018}-\left(x-5\right)^{2016}=0\)
<=> \(\left(x-5\right)^{2016}\left[\left(x-5\right)^2-1\right]=0\)
<=> \(\orbr{\begin{cases}x-5=0\\\left(x-5\right)^2=1\end{cases}}\)<=>\(\orbr{\begin{cases}x=5\\x-5=\pm1\end{cases}}\)
Vậy x\(\in\){4,5,6}
<=>
\(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\)
\(\Rightarrow\left(x-5\right)^{2016}\left[1-\left(x-5\right)^2\right]=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-5\right)^{2016}=0\\\left(x-5\right)^2=1-0=1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x-5=0\\x-5=1\\x-5=-1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=5\\x=6\\x=4\end{cases}}\)