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\(1)\)
\(VT=\left(\left|x-6\right|+\left|2022-x\right|\right)+\left|x-10\right|+\left|y-2014\right|+\left|z-2015\right|\)
\(\ge\left|x-6+2022-x\right|+\left|0\right|+\left|0\right|+\left|0\right|=2016\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\left(x-6\right)\left(2022-x\right)\ge0\left(1\right)\\x-10=y-2014=z-2015=0\left(2\right)\end{cases}}\)
\(\left(2\right)\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=10\\y=2014\\z=2015\end{cases}}\)
\(\left(1\right)\)
TH1 : \(\hept{\begin{cases}x-6\ge0\\2022-x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge6\\x\le2022\end{cases}\Leftrightarrow}6\le x\le2022}\) ( nhận )
TH2 : \(\hept{\begin{cases}x-6\le0\\2022-x\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le6\\x\ge2022\end{cases}}}\) ( loại )
Vậy \(x=10\)\(;\)\(y=2014\) và \(z=2015\)
\(2)\)
\(VT=\left|x-5\right|+\left|1-x\right|\ge\left|x-5+1-x\right|=\left|-4\right|=4\)
\(VP=\frac{12}{\left|y+1\right|+3}\le\frac{12}{3}=4\)
\(\Rightarrow\)\(VT\ge VP\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\left(x-5\right)\left(1-x\right)\ge0\left(1\right)\\\left|y+1\right|=0\left(2\right)\end{cases}}\)
\(\left(1\right)\)
TH1 : \(\hept{\begin{cases}x-5\ge0\\1-x\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ge5\\x\le1\end{cases}}}\) ( loại )
TH2 : \(\hept{\begin{cases}x-5\le0\\1-x\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le5\\x\ge1\end{cases}\Leftrightarrow}1\le x\le5}\) ( nhận )
\(\left(2\right)\)\(\Leftrightarrow\)\(y=-1\)
Vậy \(1\le x\le5\) và \(y=-1\)
Vì : (3x+1)2018+(2y-1)2018+\(\left|x+2y-z\right|\)2018=0
Nên: \(\left\{{}\begin{matrix}\left(3x+1\right)^{2018}=0\\\left(2y-1\right)^{2018}\\\left|x+2y-z\right|^{2018}=0\end{matrix}\right.=0\) ⇔\(\left\{{}\begin{matrix}3x+1=0\\2y-1=0\\x+2y-z=0\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}x=\dfrac{-1}{3}\\y=\dfrac{1}{2}\\\dfrac{-1}{3}+1-z=0\end{matrix}\right.\) ⇔\(\left\{{}\begin{matrix}x=\dfrac{-1}{3}\\y=\dfrac{1}{2}\\z=\dfrac{2}{3}\end{matrix}\right.\)
Vậy : x=\(\dfrac{-1}{3}\) ; y=\(\dfrac{1}{2}\) ; z=\(\dfrac{2}{3}\)
\(C=2\left(x-y\right)+13x^3y^2\left(x-y\right)-15xy\left(x-y\right)+1=1\)
Vậy C=1
\(C=2x-2y+13x^3y^2\left(x-y\right)+15\left(y^2x-x^2y\right)+\left(\dfrac{2015}{2016}\right)^0\)
\(C=2\left(x+y\right)+13x^3y^2\left(x-y\right)+15xy\left(x-y\right)+1\)
Mà x - y = 0 (bài cho)
\(\Rightarrow C=2.0+13x^3y^2.0+15xy.0+1\)
\(C=1\)
Vậy C=1
\(\left\{{}\begin{matrix}\left(x-1\right)^{2016}\ge0\\\left(2y-1\right)^{2016}\ge0\\\left|x+2y-z\right|^{2017}\ge0\end{matrix}\right.\Rightarrow\left(x-1\right)^{2016}+\left(2y-1\right)^{2016}+\left|x+2y-z\right|^{2017}\ge0\)
Mà \(\left(x-1\right)^{2017}+\left(2y-1\right)^{2016}+\left|x+2y-z\right|^{2017}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-1\right)^{2016}=0\\\left(2y-1\right)^{2016}=0\\\left|x+2y-z\right|^{2017}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{2}\\z=2\end{matrix}\right.\)
Thanks Nguyễn Huy Tú nhìu!!