Ta có ĐK 2016x - 2017 \(\ge\)0 

khi đó \(\orbr{\begin{cases}|6x-2|-5=2016x-2017\\|6x-2|-5=-2016x+2017\end{cases}}\)

TH1: /6x-2/ - 5 = 2016x-2017  (1)

với x\(\ge\)1/3  Từ (1) suy ra 6x-2 - 5 = 2016x - 2017

2010x = 2010 suy ra x=1 

Với x < 1/3 . Từ (1) suy ra -(6x-2) - 5 = 2016x- 2017

-6x + 2 - 5 = 2016x - 2017

2014 = 2022 x

x= 2014/2022 (loại)

TH2: /6x-2/ - 5 = -2016x+2017  (2)

các em làm  tương tự nhé

cô ơi, sao x=2014/2022 lại loại ạ ?

a) \(\left(x-\sqrt{3}\right)^2=\frac{3}{4}\)

\(\Leftrightarrow x-\sqrt{3}=\pm\frac{\sqrt{3}}{2}\)

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

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

Nghiệm cuối cùng là : \(x_1=\frac{\sqrt{3}}{2};x_2=\frac{3\sqrt{3}}{2}\)

b) || 6x - 2  | - 5 | = 2016. x -2017 

<=> || 6x - 2 | -5 | -2016x = -2017

<=> \(\orbr{\begin{cases}\left|6x-2\right|-5-2016.x=-2017,\left|6x-2\right|-5\ge0\\-\left(\left|6x-2\right|-5\right)-2016x=-2017,\left|6x-2\right|-5< 0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=1,x\in\left[-\infty,-\frac{1}{2}\right];\left[\frac{7}{6};+\infty\right]\\x=\frac{1012}{1011},x\in\left[-\frac{1}{2},\frac{7}{6}\right]\end{cases}}\)

<=>\(\orbr{\begin{cases}x\in\varnothing\\x=\frac{1012}{1011}\end{cases}}\)

Vậy x = \(\frac{1012}{1011}\)

a, Ta có:
\(\orbr{\begin{cases}x-\sqrt{\frac{3}{4}}=\sqrt{\frac{3}{4}}\\x-\sqrt{\frac{3}{4}}=-\sqrt{\frac{3}{4}}\end{cases}\Rightarrow\orbr{\begin{cases}x=2\sqrt{\frac{3}{4}}\\x=0\end{cases}}}\)

mình xin lỗi , mình ghi sai đề

a)\(\left(x-\sqrt{3}\right)^2=\frac{3}{4}\)

\(\left|\left|6x-2\right|-5\right|=2016x-2017\)

Xét trường hợp 1: \(\left|6x-2\right|-5=2016x-2017\)

\(\Rightarrow\left|6x-2\right|=2016x-2017+5\)

\(\Rightarrow\left|6x-2\right|=2016x-2012\)

\(\Rightarrow\left[{}\begin{matrix}6x-2=2016x-2012\\6x-2=-\left(2016x-2012\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}6x-2-2016x+2012=0\\6x-2+2016x-2012=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-2010x+2010=0\\2022x-2014=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}-2010x=-2010\\2022x=2014\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-2010:-2010\\x=2014:2022\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1007}{1011}\end{matrix}\right.\)

Xét trường hợp 2: \(\left|6x-2\right|-5=-\left(2016x-2017\right)\)

\(\Rightarrow\left|6x-2\right|=-\left(2016x-2017\right)+5\)

\(\Rightarrow\left|6x-2\right|=-2016x+2017+5\)

\(\Rightarrow\left|6x-2\right|=-2016x+2022\)

\(\Rightarrow\left|6x-2\right|=-\left(2016x-2022\right)\)

\(\Rightarrow\left[{}\begin{matrix}6x-2=-\left(2016x-2022\right)\\6x-2=-\left[-\left(2016x-2022\right)\right]\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}6x-2=-\left(2016x-2022\right)\\6x-2=2016x-2022\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}6x-2-\left[-\left(2016x-2022\right)\right]=0\\6x-2-\left(2016x-2022\right)=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}6x-2+2016x-2022=0\\6x-2-2016x+2022=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2022x-2024=0\\-2010x+2020=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2022x=2024\\-2010x=-2020\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2024:2022\\x=\left(-2020\right):\left(-2010\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1012}{1011}\\x=\dfrac{202}{201}\end{matrix}\right.\)

Vậy \(x=1\) hoặc \(x=\dfrac{1007}{1011}\) hoặc \(x=\dfrac{1012}{1011}\) hoặc \(x=\dfrac{202}{201}\)

tk mk đi