tìm \(x\) biết\(a)\,\,\,\left| {\frac{{ - 2}}{3}:x...

A

- Hướng dẫn giải

Phương pháp giải:

+) Sử dụng quy tắc bỏ dấu giá trị tuyệt đối: \(\left| x \right| = \left\{ \begin{array}{l}x\,\,\,\,\,khi\,\,\,x \ge 0\\- x\,\,\,\,khi\,\,\,\,x < 0\end{array} \right..\)

+) Ta có công thức \({\left( x \right)^2} = {a^2} \Leftrightarrow \left[ \begin{array}{l}x = a\\x =  - a\end{array} \right..\)

+) Đặt nhân tử chung để phương trình về dạng phương trình tích.

Giải chi tiết:

\(a)\,\,\,\left| {\frac{{ - 2}}{3}:x + \frac{1}{6}} \right| = \frac{1}{{12}}\)

TH1: Khi \( - \frac{2}{3}:x + \frac{1}{6} \ge 0 \Leftrightarrow  - \frac{2}{3}:x \ge  - \frac{1}{6} \Leftrightarrow \frac{2}{3}:x \le \frac{1}{6} \Leftrightarrow x \ge 4\) ta có: \(\left| { - \frac{2}{3}:x + \frac{1}{6}} \right| =  - \frac{2}{3}:x + \frac{1}{6}.\)

\(\begin{array}{l}\,\,\,\,\,\,\,\left| { - \frac{2}{3}:x + \frac{1}{6}} \right| = \frac{1}{{12}}\\\Leftrightarrow  - \frac{2}{3}:x + \frac{1}{6} = \frac{1}{{12}}\\\Leftrightarrow  - \frac{2}{3}:x =  - \frac{1}{{12}}\\\Leftrightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,x =  - \frac{2}{3}:\left( { - \frac{1}{{12}}} \right)\\\Leftrightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,x = 8\,\,\,\,\left( {tm} \right)\end{array}\)

TH2: \( - \frac{2}{3}:x + \frac{1}{6} \le 0 \Leftrightarrow  - \frac{2}{3}:x \le  - \frac{1}{6} \Leftrightarrow \frac{2}{3}:x \ge \frac{1}{6} \Leftrightarrow x \le 4\)ta có  \(\left| { - \frac{2}{3}:x + \frac{1}{6}} \right| = \frac{2}{3}:x - \frac{1}{6}.\)

\(\begin{array}{l}\,\,\,\,\,\,\,\left| { - \frac{2}{3}:x + \frac{1}{6}} \right| = \frac{1}{{12}}\\\Leftrightarrow \frac{2}{3}:x - \frac{1}{6} = \frac{1}{{12}}\\ \Leftrightarrow \frac{2}{3}:x\,\,\,\,\,\,\,\,\,\, = \frac{1}{{12}} + \frac{1}{6}\\ \Leftrightarrow \,\frac{2}{3}:x\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{4}\\\Leftrightarrow \,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\, = \frac{2}{3}:\frac{1}{4}\\\Leftrightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,x = \frac{8}{3}\,\,\,\,\left( {tm} \right)\end{array}\)

Vậy \(x = 8\) hoặc \(x = \frac{8}{3}.\)

\(\begin{array}{l}b)\,\,3.\left( {x + \frac{1}{2}} \right) - \frac{{{x^2}}}{3}\left( {x + \frac{1}{2}} \right) = 0\\ \Leftrightarrow \left( {x + \frac{1}{2}} \right)\left( {3 - \frac{{{x^2}}}{3}} \right) = 0\\\Leftrightarrow \left[ \begin{array}{l}x + \frac{1}{2} = 0\\3 - \frac{{{x^2}}}{3} = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x =  - \frac{1}{2}\\{x^2} = 9\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x =  - \frac{1}{2}\\x =  \pm 3\end{array} \right..\end{array}\) 

Vậy \(x \in \left\{ { - 3; - \frac{1}{2};\,\,3} \right\}.\)

\(\begin{array}{l}c)\,\,\,{\left( {x - \sqrt 3 } \right)^2} = \frac{3}{4}\\\Leftrightarrow \left[ \begin{array}{l}x - \sqrt 3  = \sqrt {\frac{3}{4}} \\x - \sqrt 3  =  - \sqrt {\frac{3}{4}}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x - \sqrt 3  = \frac{{\sqrt 3 }}{2}\\x - \sqrt 3  =  - \frac{{\sqrt 3 }}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{{3\sqrt 3 }}{2}\\x = \frac{{\sqrt 3 }}{2}\end{array} \right..\end{array}\)  

Vậy \(x \in \left\{ {\frac{{3\sqrt 3 }}{2};\frac{{\sqrt 3 }}{2}} \right\}.\) 

\(\begin{array}{l}d)\left| {\left| {6x - 2} \right| - 5} \right| = 2016x - 2017\\TH1:\left| {6x - 2} \right| \ge 5 \Leftrightarrow \left[ \begin{array}{l}6x - 2 \ge 5\\6x - 2 \le - 5\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x \ge \frac{7}{6}\\x \le - \frac{1}{2}\end{array} \right..\\ \Rightarrow \left| {\left| {6x - 2} \right| - 5} \right| = \left| {6x - 2} \right| - 5\\ \Rightarrow pt \Leftrightarrow \left| {6x - 2} \right| - 5 = 2016x - 2017\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \left| {6x - 2} \right| = 2016x - 2012\end{array}\)

+) Với  \(x \ge \frac{7}{6}\) ta có \(\left| {6x - 2} \right| = 6x - 2\):

\(\begin{array}{l}\Rightarrow pt \Leftrightarrow 6x - 2 - 5 = 2016x - 2017\\\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow 2010x\,\,\,\,\,\,\,\,\,\, = 2010\\\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1\,\,\,\,\,\left( {ktm} \right).\end{array}\)

+) Với \(x \le  - \frac{1}{2} \Rightarrow \left| {6x - 2} \right| = 2 - 6x.\)

\(\begin{array}{l} \Rightarrow pt \Leftrightarrow 2 - 6x - 5 = 2016x - 2017\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow 2022x\,\,\,\,\,\,\,\,\, = 2014\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{1007}}{{1011}}\,\,\,\left( {ktm} \right).\end{array}\)

\(\begin{array}{l}TH2:\,\,\,\left| {6x - 2} \right| < 5 \Leftrightarrow  - 5 < 6x - 2 < 5\\ \Leftrightarrow  - 3 < 6x < 7 \Leftrightarrow  - \frac{1}{2} < x < \frac{7}{6}.\\ \Rightarrow \left| {\left| {6x - 2} \right| - 5} \right| = 5 - \left| {6x - 2} \right|\\\Leftrightarrow 5 - \left| {6x - 2} \right| = 2016x - 2017\\\Leftrightarrow \left| {6x - 2} \right| = 2022 - 2016x\end{array}\)

+) Với \(6x - 2 \ge 0 \Leftrightarrow x \ge \frac{1}{3} \Rightarrow \left| {6x - 2} \right| = 6x - 2.\) Xét \(\frac{1}{3} \le x < \frac{7}{6}\) ta có:

\(\begin{array}{l}pt \Leftrightarrow 6x - 2 = 2022 - 2016x\\\,\,\,\,\,\, \Leftrightarrow 2022x\,\,\, = 2024\\\,\,\,\,\,\, \Leftrightarrow x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{1012}}{{1011}}\,\,\,\left( {tm} \right)\end{array}\)

+) Với \(6x - 2 < 0 \Leftrightarrow x < \frac{1}{3} \Rightarrow \left| {6x - 2} \right| = 2 - 6x.\) Xét \( - \frac{1}{2} < x < \frac{1}{3}\) ta có:

\(\begin{array}{l}pt \Leftrightarrow 2 - 6x = 2022 - 2016x\\\,\,\,\,\,\, \Leftrightarrow 2010x = 2020\\\,\,\,\,\,\, \Leftrightarrow x\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{202}}{{201}}\,\,\,\,\left( {ktm} \right).\end{array}\)

Vậy \(x = \frac{{1012}}{{1011}}.\)

Chọn đáp án A.