Vì |x| + 2017 \(\ge2017>0\forall x\)

=> 504|x| - 2016 < 0

=> 504|x| < 2016

=> |x| < 4

=> -4 < x < 4

=> \(x\in\left\{-3;-2;-1;0;1;2;3\right\}\)(Vì x nguyên)

Ta có: ||3x-3|+2x+\(\left(-1\right)^{2016}\) |=3x+\(2017^0\)

=>||3x-3|+2x+1|=3x+1(1)

TH1: |3x-3|+2x+1=3x+1

=>|3x-3|=3x+1-2x-1=x

=>\(\begin{cases}x\ge0\\ x^2=\left(3x-3\right)^2\end{cases}\Rightarrow\begin{cases}x\ge0\\ \left(3x-3-x\right)\left(3x-3+x\right)=0\end{cases}\)

=>\(\begin{cases}x\ge0\\ \left(2x-3\right)\left(4x-3\right)=0\end{cases}\Rightarrow x\in\left\lbrace\frac32;\frac34\right\rbrace\)

THay lại vào trong (1), ta thấy cả x=3/2 và x=3/4 đều thỏa mãn

TH2: |3x-3|+2x+1=-3x-1

=>|3x-3|=-3x-1-2x-1

=>|3x-3|=-5x-2

=>\(\begin{cases}-5x-2\ge0\\ \left(-5x-2\right)^2=\left(3x-3\right)^2\end{cases}\Rightarrow\begin{cases}-5x\ge2\\ \left(5x+2\right)^2-\left(3x-3\right)^2=0\end{cases}\)

=>\(\begin{cases}x\le-\frac25\\ \left(5x+2-3x+3\right)\left(5x+2+3x-3\right)=0\end{cases}\Rightarrow\begin{cases}x\le-\frac25\\ \left(2x+5\right)\left(8x-1\right)=0\end{cases}\)

=>\(x=-\frac52\)

Thay lại vào (1), ta thấy x=-5/2 không thỏa mãn

=>Loại

Vậy: \(x\in\left\lbrace\frac32;\frac34\right\rbrace\)

|x-2016|²⁰¹⁶+|x-2017|²⁰¹⁶=1

|x-2016|²⁰¹⁶=1 hoặc |x-2017|²⁰¹⁶=1

th1:|x-2016|²⁰¹⁶=1                                                    

|x-2016|²⁰¹⁶=1²⁰¹⁶                                                                       

x-2016=1

x=1+2016

x=2017 

th2:

làm tương tự

x=2016hoac x=2017

\(a)\) \(\left|\left|3x-3\right|2x+\left(-1\right)^{2016}\right|=3x+2017^0\)

\(\Leftrightarrow\)\(\left|\left|3x-3\right|2x+1\right|=3x+1\)

Mà \(\left|\left|3x-3\right|2x+1\right|\ge0\) nên \(3x+1\ge0\)\(\Rightarrow\)\(x\ge1\)

\(\Leftrightarrow\)\(\left|3x-3\right|2x+1=3x+1\)

\(\Leftrightarrow\)\(\left|3x-3\right|=\frac{3x}{2x}\)

\(\Leftrightarrow\)\(\left|3x-3\right|=\frac{3}{2}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x-3=\frac{3}{2}\\3x-3=\frac{-3}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=\frac{9}{2}\\3x=\frac{3}{2}\end{cases}}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{9}{2}:3\\x=\frac{3}{2}:3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\left(tmx\ge1\right)\\x=\frac{1}{2}\left(loai\right)\end{cases}}}\)

Vậy \(x=\frac{3}{2}\)

\(1)\)

\(VT=\left(\left|x-6\right|+\left|2022-x\right|\right)+\left|x-10\right|+\left|y-2014\right|+\left|z-2015\right|\)

\(\ge\left|x-6+2022-x\right|+\left|0\right|+\left|0\right|+\left|0\right|=2016\)

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 \(y=-1\)