Bài 3.2:

a, ||2x - 1| + $\frac{1}{2}$| = $\frac{3}{4}$

⇔ \(\left[ \begin{array}{l}|2x-1|+\frac{1}{2}=\frac{3}{4}\\|2x-1|+\frac{1}{2}=-\frac{3}{4}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}|2x-1|=\frac{1}{4}\\|2x-1|=-\frac{5}{4}(loại)\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}2x-1=\frac{1}{4}\\2x-1=-\frac{1}{4}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}2x=\frac{5}{4}\\2x=\frac{3}{4}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=\frac{5}{8}\\x=\frac{3}{8}\end{array} \right.\) 

b, |x² + 3|x - $\frac{1}{2}$|| = x² + 3          

⇔ \(\left[ \begin{array}{l}x²+3|x-\frac{1}{2}|=x²+3\\x²+3|x-\frac{1}{2}|=-x²-3\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}3|x-\frac{1}{2}|=3\\3|x-\frac{1}{2}|=-2x²-3(loại)\end{array} \right.\) 

⇔ |x - $\frac{1}{2}$| = 1

⇔ \(\left[ \begin{array}{l}x-\frac{1}{2}=1\\x-\frac{1}{2}=-1\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=\frac{3}{2}\\x=-\frac{1}{2}\end{array} \right.\) 

Bài 3.3:

a, |$\frac{3}{2}$x + $\frac{1}{2}$| = |4x - 1|

⇔ \(\left[ \begin{array}{l}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=\frac{3}{5}\\x=\frac{1}{11}\end{array} \right.\) 

b, |$\frac{7}{5}$x + $\frac{1}{2}$| = |$\frac{4}{3}$x - $\frac{1}{4}$|

⇔ \(\left[ \begin{array}{l}\frac{7}{5}x+\frac{1}{2}=\frac{4}{3}x-\frac{1}{4}\\\frac{7}{5}x+\frac{1}{2}=\frac{1}{4}-\frac{4}{3}x\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}\frac{1}{15}x=-\frac{3}{4}\\\frac{41}{15}x=-\frac{1}{4}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=-\frac{45}{4}\\x=-\frac{15}{164}\end{array} \right.\) 

c, |$\frac{5}{4}$x - $\frac{7}{2}$| - |$\frac{5}{8}$x + $\frac{3}{5}$| = 0

⇔ |$\frac{5}{4}$x - $\frac{7}{2}$| = |$\frac{5}{8}$x + $\frac{3}{5}$| 

⇔ \(\left[ \begin{array}{l}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}-\frac{3}{5}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=\frac{164}{25}\\x=\frac{116}{75}\end{array} \right.\) 

d, |$\frac{7}{8}$x + $\frac{5}{6}$| - |$\frac{1}{2}$x + 5| = 0

⇔ |$\frac{7}{8}$x + $\frac{5}{6}$| = |$\frac{1}{2}$x + 5| 

⇔ \(\left[ \begin{array}{l}\frac{7}{8}x+\frac{5}{6}=\frac{1}{2}x+5\\\frac{7}{8}x+\frac{5}{6}=-\frac{1}{2}x-5\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}\frac{3}{8}x=\frac{25}{6}\\\frac{11}{8}x=-\frac{35}{6}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=\frac{100}{9}\\x=-\frac{140}{33}\end{array} \right.\) 

Bài 3.4:

a, |5 - $\frac{2}{3}$x| + |$\frac{2}{3}$y - 4| = 0

Vì |5 - $\frac{2}{3}$x| ≥ 0; |$\frac{2}{3}$y - 4| ≥ 0 

mà |5 - $\frac{2}{3}$x| + |$\frac{2}{3}$y - 4| = 0

⇒ $\left \{ {{|5-\frac{2}{3}x|=0} \atop {|\frac{2}{3}y-4|=0}} \right.$ 

⇔ $\left \{ {{5-\frac{2}{3}x=0} \atop {\frac{2}{3}y-4=0}} \right.$ 

⇔ $\left \{ {{\frac{2}{3}x=5} \atop {\frac{2}{3}y=4}} \right.$ 

⇔ $\left \{ {{x=\frac{15}{2}} \atop {y=6}} \right.$ 

b, |$\frac{2}{3}$ - $\frac{1}{2}$ + $\frac{3}{4}$x| + |1,5 - $\frac{3}{4}$ - $\frac{3}{2}$y| = 0

⇔ |$\frac{1}{6}$ + $\frac{3}{4}$x| + |$\frac{3}{4}$ - $\frac{3}{2}$y| = 0

Vì |$\frac{1}{6}$ + $\frac{3}{4}$x| ≥ 0; |$\frac{3}{4}$ - $\frac{3}{2}$y| ≥ 0

mà |$\frac{1}{6}$ + $\frac{3}{4}$x| + |$\frac{3}{4}$ - $\frac{3}{2}$y| = 0

⇒ $\left \{ {{|\frac{1}{6}+\frac{3}{4}x|=0} \atop {|\frac{3}{4}-\frac{3}{2}y|=0}} \right.$ 

⇔ $\left \{ {{\frac{1}{6}+\frac{3}{4}x=0} \atop {\frac{3}{4}-\frac{3}{2}y=0}} \right.$ 

⇔  $\left \{ {{\frac{3}{4}x=-\frac{1}{6}} \atop {\frac{3}{2}y=\frac{3}{4}}} \right.$ 

⇔ $\left \{ {{x=-\frac{2}{9}} \atop {y=\frac{1}{2}}} \right.$ 

c, |x - 2020| + |y - 2021| = 0

Vì |x - 2020| ≥ 0; |y - 2021| ≥ 0

mà |x - 2020| + |y - 2021| = 0

⇒ $\left \{ {{|x-2020|=0} \atop {|y-2021|=0}} \right.$ 

⇔ $\left \{ {{x-2020=0} \atop {y-2021=0}} \right.$ 

⇔ $\left \{ {{x=2020} \atop {y=2021}} \right.$ 

d, |x - y| + |y + $\frac{21}{10}$| = 0

Vì |x - y| ≥ 0; |y + $\frac{21}{10}$| ≥ 0

mà |x - y| + |y + $\frac{21}{10}$| = 0

⇒ $\left \{ {{|x-y|=0} \atop {|y+\frac{21}{10}|=0}} \right.$ 

⇔ $\left \{ {{x-y=0} \atop {y+\frac{21}{10}=0}} \right.$ 

⇔ $\left \{ {{x=y} \atop {y=-\frac{21}{10}}} \right.$ 

⇒ x = y = -$\frac{21}{10}$

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Bài 3.2:

a, ||2x - 1| + $\frac{1}{2}$| = $\frac{3}{4}$

⇔\(\left[\begin{array}{l}  |2x-1|+\frac{1}{2}=\frac{3}{4}\\|2x-1|+\frac{1}{2}=-\frac{3}{4}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}|2x-1|=\frac{1}{4}\\|2x-1|=-\frac{5}{4}(loại)\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}2x-1=\frac{1}{4}\\2x-1=-\frac{1}{4}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}2x=\frac{5}{4}\\2x=\frac{3}{4}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=\frac{5}{8}\\x=\frac{3}{8}\end{array} \right.\) 

b, |x² + 3|x - $\frac{1}{2}$|| = x² + 3          

⇔ \(\left[ \begin{array}{l}x²+3|x-\frac{1}{2}|=x²+3\\x²+3|x-\frac{1}{2}|=-x²-3\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}3|x-\frac{1}{2}|=3\\3|x-\frac{1}{2}|=-2x²-3(loại)\end{array} \right.\) 

⇔ |x - $\frac{1}{2}$| = 1

⇔ \(\left[ \begin{array}{l}x-\frac{1}{2}=1\\x-\frac{1}{2}=-1\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=\frac{3}{2}\\x=-\frac{1}{2}\end{array} \right.\) 

Bài 3.3:

a, |$\frac{3}{2}$x + $\frac{1}{2}$| = |4x - 1|

⇔ \(\left[ \begin{array}{l}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=\frac{3}{5}\\x=\frac{1}{11}\end{array} \right.\) 

b, |$\frac{7}{5}$x + $\frac{1}{2}$| = |$\frac{4}{3}$x - $\frac{1}{4}$|

⇔ \(\left[ \begin{array}{l}\frac{7}{5}x+\frac{1}{2}=\frac{4}{3}x-\frac{1}{4}\\\frac{7}{5}x+\frac{1}{2}=\frac{1}{4}-\frac{4}{3}x\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}\frac{1}{15}x=-\frac{3}{4}\\\frac{41}{15}x=-\frac{1}{4}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=-\frac{45}{4}\\x=-\frac{15}{164}\end{array} \right.\) 

c, |$\frac{5}{4}$x - $\frac{7}{2}$| - |$\frac{5}{8}$x + $\frac{3}{5}$| = 0

⇔ |$\frac{5}{4}$x - $\frac{7}{2}$| = |$\frac{5}{8}$x + $\frac{3}{5}$| 

⇔ \(\left[ \begin{array}{l}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}-\frac{3}{5}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=\frac{164}{25}\\x=\frac{116}{75}\end{array} \right.\) 

d, |$\frac{7}{8}$x + $\frac{5}{6}$| - |$\frac{1}{2}$x + 5| = 0

⇔ |$\frac{7}{8}$x + $\frac{5}{6}$| = |$\frac{1}{2}$x + 5| 

⇔ \(\left[ \begin{array}{l}\frac{7}{8}x+\frac{5}{6}=\frac{1}{2}x+5\\\frac{7}{8}x+\frac{5}{6}=-\frac{1}{2}x-5\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}\frac{3}{8}x=\frac{25}{6}\\\frac{11}{8}x=-\frac{35}{6}\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x=\frac{100}{9}\\x=-\frac{140}{33}\end{array} \right.\) 

Bài 3.4:

a, |5 - $\frac{2}{3}$x| + |$\frac{2}{3}$y - 4| = 0

Vì |5 - $\frac{2}{3}$x| ≥ 0; |$\frac{2}{3}$y - 4| ≥ 0 

mà |5 - $\frac{2}{3}$x| + |$\frac{2}{3}$y - 4| = 0

⇒ $\left \{ {{|5-\frac{2}{3}x|=0} \atop {|\frac{2}{3}y-4|=0}} \right.$ 

⇔ $\left \{ {{5-\frac{2}{3}x=0} \atop {\frac{2}{3}y-4=0}} \right.$ 

⇔ $\left \{ {{\frac{2}{3}x=5} \atop {\frac{2}{3}y=4}} \right.$ 

⇔ $\left \{ {{x=\frac{15}{2}} \atop {y=6}} \right.$ 

b, |$\frac{2}{3}$ - $\frac{1}{2}$ + $\frac{3}{4}$x| + |1,5 - $\frac{3}{4}$ - $\frac{3}{2}$y| = 0

⇔ |$\frac{1}{6}$ + $\frac{3}{4}$x| + |$\frac{3}{4}$ - $\frac{3}{2}$y| = 0

Vì |$\frac{1}{6}$ + $\frac{3}{4}$x| ≥ 0; |$\frac{3}{4}$ - $\frac{3}{2}$y| ≥ 0

mà |$\frac{1}{6}$ + $\frac{3}{4}$x| + |$\frac{3}{4}$ - $\frac{3}{2}$y| = 0

⇒ $\left \{ {{|\frac{1}{6}+\frac{3}{4}x|=0} \atop {|\frac{3}{4}-\frac{3}{2}y|=0}} \right.$ 

⇔ $\left \{ {{\frac{1}{6}+\frac{3}{4}x=0} \atop {\frac{3}{4}-\frac{3}{2}y=0}} \right.$ 

⇔  $\left \{ {{\frac{3}{4}x=-\frac{1}{6}} \atop {\frac{3}{2}y=\frac{3}{4}}} \right.$ 

⇔ $\left \{ {{x=-\frac{2}{9}} \atop {y=\frac{1}{2}}} \right.$ 

c, |x - 2020| + |y - 2021| = 0

Vì |x - 2020| ≥ 0; |y - 2021| ≥ 0

mà |x - 2020| + |y - 2021| = 0

⇒ $\left \{ {{|x-2020|=0} \atop {|y-2021|=0}} \right.$ 

⇔ $\left \{ {{x-2020=0} \atop {y-2021=0}} \right.$ 

⇔ $\left \{ {{x=2020} \atop {y=2021}} \right.$ 

d, |x - y| + |y + $\frac{21}{10}$| = 0

Vì |x - y| ≥ 0; |y + $\frac{21}{10}$| ≥ 0

mà |x - y| + |y + $\frac{21}{10}$| = 0

⇒ $\left \{ {{|x-y|=0} \atop {|y+\frac{21}{10}|=0}} \right.$ 

⇔ $\left \{ {{x-y=0} \atop {y+\frac{21}{10}=0}} \right.$ 

⇔ $\left \{ {{x=y} \atop {y=-\frac{21}{10}}} \right.$ 

⇒ x = y = -$\frac{21}{10}$