`2.1`

`a)|5x-4|=|x+2|`

`⇒`\(\left[ \begin{array}{l}5x-4=x+2\\5x-4=-x-2\end{array} \right.\) 

`⇒`\(\left[ \begin{array}{l}5x-x=2+4\\5x+x=-2+4\end{array} \right.\) 

`⇒`\(\left[ \begin{array}{l}4x=6\\6x=2\end{array} \right.\) 

`⇒`\(\left[ \begin{array}{l}x=\dfrac{6}{4}\\x=\dfrac{2}{6}\end{array} \right.\) 

`⇒`\(\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{array} \right.\) 

Vậy `x=3/2` hoặc `x=1/3`

`b)|2x-3|-|3x+2|=0`

   `|2x-3|=|3x+2|`

`⇒`\(\left[ \begin{array}{l}2x-3=3x+2\\2x-3=-3x-2\end{array} \right.\) 

`⇒`\(\left[ \begin{array}{l}2x-3x=2+3\\2x+3x=-2+3\end{array} \right.\) 

`⇒`\(\left[ \begin{array}{l}-x=5\\5x=1\end{array} \right.\) 

`⇒`\(\left[ \begin{array}{l}x=-5\\x=\dfrac{1}{5}\end{array} \right.\) 

Vậy `x=-5` hoặc `x=1/5`

`c)|2+3x|=|4x-3|`

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

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

`⇒`\(\left[ \begin{array}{l}-x=-5\\7x=1\end{array} \right.\) 

`⇒`\(\left[ \begin{array}{l}x=5\\x=\dfrac{1}{7}\end{array} \right.\) 

Vậy `x=5` hoặc `x=1/7`

`d)|7x+1|-|5x+6|=0`

   `|7x+1|=|5x+6|`

`⇒`\(\left[ \begin{array}{l}7x+1=5x+6\\7x+1=-5x-6\end{array} \right.\) 

`⇒`\(\left[ \begin{array}{l}7x-5x=6-1\\7x+5x=-6-1\end{array} \right.\) 

`⇒`\(\left[ \begin{array}{l}2x=5\\12x=-7\end{array} \right.\) 

`⇒`\(\left[ \begin{array}{l}x=\dfrac{5}{2}\\x=\dfrac{-7}{12}\end{array} \right.\) 

Vậy `x=5/2` hoặc `x=-7/(12)`

`2.2`

`a)|3/2x+1/2|=|4x-1|`

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

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

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

`⇒`\(\left[ \begin{array}{l}x=\dfrac{3}{5}\\x=\dfrac{1}{11}\end{array} \right.\) 

Vậy `x=3/5` hoặc `x=1/(11)`

`b)|5/4x-7/2|-|5/8x+3/5|=0`

   `|5/4x-7/2|=|5/8x+3/5|`

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

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

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

`⇒`\(\left[ \begin{array}{l}x= \dfrac{164}{25}\\x= \dfrac{116}{75}\end{array} \right.\) 

Vậy `x=(164)/(25)` hoặc `x=(116)/(75)`

`c)|7/5x+2/3|=|4/3x-1/4|`

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

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

`⇒`\(\left[ \begin{array}{l} \dfrac{1}{15}x= \dfrac{-11}{12}\\ \dfrac{41}{15}x= \dfrac{-5}{12}\end{array} \right.\) 

`⇒`\(\left[ \begin{array}{l}x= \dfrac{-55}{4}\\x= \dfrac{-25}{164}\end{array} \right.\) 

Vậy `x=(-55)/4` hoặc `x=(-25)/(164)`

`d)|7/8x+5/6|-|1/2x+5|=0`

   `|7/8x+5/6|=|1/2x+5|`

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

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

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

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

Vậy `x=(100)/9` hoặc `x=(-140)/(33)`

Đáp án:

Bài `2.1`

`a,`

`|5x-4|=|x+2|`

`->` \(\left[ \begin{array}{l}5x-4=x+2\\5x-4=-x-2\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}5x-x=4+2\\5x+x=4-2\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}4x=6\\6x=2\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{array} \right.\) 

Vậy `x=3/2` hoặc `x=1/3`

`b,`

`|2x-3| - |3x+2|=0`

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

`->` \(\left[ \begin{array}{l}2x-3=3x+2\\2x-3=-3x-2\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}2x-3x=3+2\\2x+3x=3-2\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}-x=5\\5x=1\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}x=-5\\x=\dfrac{1}{5}\end{array} \right.\) 

Vậy `x=-5` hoặc `x=1/5`

`c,`

`|2 + 3x|= |4x - 3|`

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

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

`->` \(\left[ \begin{array}{l}-x=-5\\7x=1\end{array} \right.\) 

`->`\(\left[ \begin{array}{l}x=5\\x=\dfrac{1}{7}\end{array} \right.\) 

Vậy `x=5` hoặc `x=1/7`

`d,`

`|7x + 1| - |5x + 6| = 0`

`-> |7x + 1| = |5x + 6|`

`->` \(\left[ \begin{array}{l}7x+1=5x+6\\7x+1=-5x-6\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}7x-5x=-1+6\\7x+5x=-1-6\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}2x=5\\12x=-7\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}x=\dfrac{5}{2}\\x=\dfrac{-7}{12}\end{array} \right.\) 

Vậy `x=5/2` hoặc `x=(-7)/12`

Bài `2.2`

`a,`

`|3/2x + 1/2| = |4x - 1|`

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

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

`->` \(\left[ \begin{array}{l}x=\dfrac{3}{5}\\x=\dfrac{1}{11}\end{array} \right.\) 

Vậy `x=3/5` hoặc `x=1/11`

`b,`

`|5/4x - 7/2| - |5/8x + 3/5| =0`

`->|5/4x-7/2|=|5/8x+3/5|`

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

`->` \(\left[ \begin{array}{l}\dfrac{5}{8}x=\dfrac{41}{10}\\ \dfrac{15}{8}x=\dfrac{29}{10}\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}x=\dfrac{164}{25}\\x=\dfrac{116}{75}\end{array} \right.\) 

Vậy `x=164/25` hoặc `x=116/75`

`c,`

`|7/5x + 2/3| = |4/3x - 1/4|`

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

`->` \(\left[ \begin{array}{l}\dfrac{1}{15}x=\dfrac{-11}{12}\\ \dfrac{41}{15}x=\dfrac{-5}{12}\end{array} \right.\) 

`->` \(\left[ \begin{array}{l}x=\dfrac{-55}{4}\\x=\dfrac{-25}{164}\end{array} \right.\) 

Vậy `x=(-55)/4` hoặc `x=(-25)/164`

`d,`

`|7/8x + 5/6| - |1/2x + 5| = 0`

`-> |7/8x + 5/6| = |1/2x + 5|`

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

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

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

Vậy `x=100/9` hoặc `x=(-140)/33`