Tìm x theo 2 cách : a) 2/5 - | x + 1 | = 3/4 B) | x - 1/3 | = 2/3 C) 4 | x - 1/2 | = 1/3

a) \(\dfrac{2}{5} - | x + 1 | = \dfrac{3}{4}\)

TH1: \(x+1<0\Leftrightarrow x< -1\)

pt \(\Rightarrow \dfrac{2}{5}-[-(x+1)]=\dfrac{3}{4}\)

\(\Rightarrow \dfrac{2}{5}+x+1=\dfrac{3}{4}\)

\(\Rightarrow x+\left({\dfrac{2}{5}+1}\right)=\dfrac{3}{4}\)

\(\Rightarrow x+\dfrac{2+5}{5}=\dfrac{3}{4}\)

\(\Rightarrow x+\dfrac{7}{5}=\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{3}{4}-\dfrac{7}{5}\)

\(\Rightarrow x=\dfrac{3.5-7.4}{20}\)

\(\Rightarrow x=\dfrac{-13}{20}\) (loại).

TH2: \(x+1\ge0\Leftrightarrow x\ge -1\)

pt \(\Rightarrow \dfrac{2}{5}-(x+1)=\dfrac{3}{4}\)

\(\Rightarrow \dfrac{2}{5}-x-1=\dfrac{3}{4}\)

\(\Rightarrow \left({\dfrac{2}{5}-1}\right)-x=\dfrac{3}{4}\)

\(\Rightarrow \dfrac{2-5}{5}-x=\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{-3}{5}-\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{-3.4-3.5}{20}\)

\(\Rightarrow x=\dfrac{-27}{20}\) (loại).

b) \(| x - \dfrac{1}{3} | = \dfrac{2}{3}\)

c) \(4 | x - \dfrac{1}{2} | = \dfrac{1}{3}\)

Đáp án:

Giải thích các bước giải: \(\begin{array}{l} a)\,\,\,\frac{2}{5} - \left| {x + 1} \right| = \frac{3}{4}\\ & \left| {x + 1} \right| = \frac{2}{5} - \frac{3}{4}\\ & \left| {x + 1} \right| = \frac{{ - 7}}{{20}}\,\,\,\,(vo\,\,li)\\ b)\left| {x - \frac{1}{3}} \right| = \frac{2}{3}\\ \Rightarrow \left[ \begin{array}{l} x - \frac{1}{3} = \frac{2}{3}\\ x - \frac{1}{3} = \frac{{ - 2}}{3} \end{array} \right. \Rightarrow \left[ \begin{array}{l} x = \frac{2}{3} + \frac{1}{3}\\ x = \frac{{ - 2}}{3} + \frac{1}{3} \end{array} \right. \Rightarrow \left[ \begin{array}{l} x = 1\\ x = \frac{{ - 1}}{3} \end{array} \right.\\ c)\,\,\,4\left| {x - \frac{1}{2}} \right| = \frac{1}{3}\\ \,\,\,\,\,\,\,\,\,\,\,\left| {x - \frac{1}{2}} \right| = \frac{1}{3}:4\\ \,\,\,\,\,\,\,\,\,\,\left| {x - \frac{1}{2}} \right| = \frac{1}{{12}}\\ \Rightarrow \left[ \begin{array}{l} x - \frac{1}{2} = \frac{1}{{12}}\\ x - \frac{1}{2} = \frac{{ - 1}}{{12}} \end{array} \right. \Rightarrow \left[ \begin{array}{l} x = \frac{1}{{12}} + \frac{1}{2}\\ x = \frac{{ - 1}}{{12}} + \frac{1}{2} \end{array} \right. \Rightarrow \left[ \begin{array}{l} x = \frac{7}{{12}}\\ x = \frac{5}{{12}} \end{array} \right. \end{array}\)