\(\left(x-\frac{1}{5}\right)^{2014}+\left(y+0,4\right)^{2016}+\left(z-3\right)^{2018}=0\)

Ta thấy: \(\begin{cases}\left(x-\frac{1}{5}\right)^{2014}\ge0\\\left(y+0,4\right)^{2016}\ge0\\\left(z-3\right)^{2018}\ge0\end{cases}\)

\(\Rightarrow\left(x-\frac{1}{5}\right)^{2014}+\left(y+0,4\right)^{2016}+\left(z-3\right)^{2018}\ge0\)

\(\Rightarrow\begin{cases}\left(x-\frac{1}{5}\right)^{2014}=0\\\left(y+0,4\right)^{2016}=0\\\left(z-3\right)^{2018}=0\end{cases}\)\(\Rightarrow\begin{cases}x-\frac{1}{5}=0\\y+0,4=0\\z-3=0\end{cases}\)\(\Rightarrow\begin{cases}x=\frac{1}{5}\\y=-0,4\\z=3\end{cases}\)

lần sau viết đề cẩn thận hơn nhé

Dễ thấy: \(\left\{{}\begin{matrix}\left|x+2016\right|\ge0\\\left|x+2017\right|\ge0\\\left|x+2018\right|\ge0\end{matrix}\right.\)\(\forall x\)

\(\Rightarrow\left|x+2016\right|+\left|x+2017\right|+\left|x+2018\right|\ge0\forall x\)

\(\Rightarrow VT\ge0\Rightarrow VP\ge0\Rightarrow10x\ge0\Rightarrow x\ge10\)

\(pt\Leftrightarrow\left(x+2016\right)+\left(x+2017\right)+\left(x+2018\right)=10x\)

\(\Leftrightarrow3x+6051=10x\)

\(\Leftrightarrow6051=7x\Rightarrow x=\dfrac{6051}{7}\)

Sao chép

\(\left(2x-5\right)^{2016}+\left(3y+4\right)^{2018}\le0\)

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!

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 \(\left|x+2016\right|+\left|x+2017\right|+\left|x+2017\right|=2\Leftrightarrow x=-2017\)

ko ai biết làm à