10) $|x| = -2\dfrac{1}{3}$
$\text{ <=> vô nghiệm vì |x| ≥ 0 với mọi x }$
.
11) $|x-5|= \dfrac{1}{4}$
<=> \(\left[ \begin{array}{l}x-5=\dfrac{1}{4}\\x-5=-\dfrac{1}{4}\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=\dfrac{1}{4}+5\\x=-\dfrac{1}{4}+5\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=\dfrac{1}{4}+\dfrac{20}{4}\\x=-\dfrac{1}{4}+\dfrac{20}{4}\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=\dfrac{1+20}{4}\\x=\dfrac{-1+20}{4}\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=\dfrac{21}{4}\\x=\dfrac{19}{4}\end{array} \right.\)
.
12) $|x+3|=1\dfrac{1}{4}$
<=> \(\left[ \begin{array}{l}x+3=1\dfrac{1}{4}\\x+3=-1\dfrac{1}{4}\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x+3=\dfrac{5}{4}\\x+3=-\dfrac{5}{4}\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=\dfrac{5}{4}-3\\x=-\dfrac{5}{4}-3\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=\dfrac{5}{4}-\dfrac{12}{4}\\x=-\dfrac{5}{4}-\dfrac{12}{4}\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=\dfrac{5-12}{4}\\x=\dfrac{-5-12}{4}\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=\dfrac{-7}{4}\\x=\dfrac{-17}{4}\end{array} \right.\)
.
13) $|x-7|= -\dfrac{-5}{3}=\dfrac{5}{3}$
<=> \(\left[ \begin{array}{l}x-7=\dfrac{5}{3}\\x-7=-\dfrac{5}{3}\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=\dfrac{5}{3}+7\\x-7=-\dfrac{5}{3}+7\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=\dfrac{5}{3}+\dfrac{21}{3}\\x=-\dfrac{5}{3}+\dfrac{21}{3}\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=\dfrac{5+21}{3}\\x=\dfrac{-5+21}{3}\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=\dfrac{26}{3}\\x=\dfrac{16}{3}\end{array} \right.\)
.
14) $|x+9|= -\dfrac{4}{-3}= \dfrac{4}{3} $
<=> \(\left[ \begin{array}{l}x+9=\dfrac{4}{3}\\x+9=-\dfrac{4}{3}\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=\dfrac{4}{3}-9\\x=-\dfrac{4}{3}-9\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=\dfrac{4}{3}-\dfrac{27}{3}\\x=-\dfrac{4}{3}-\dfrac{27}{3}\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=\dfrac{4-27}{3}\\x=\dfrac{-4-27}{3}\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=\dfrac{-23}{3}\\x=\dfrac{-31}{3}\end{array} \right.\)
.
15) $|x-1|= 3\dfrac{1}{2}=\dfrac{7}{2}$
<=> \(\left[ \begin{array}{l}x-1=\dfrac{7}{2}\\x-1=-\dfrac{7}{2}\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=\dfrac{7}{2}+1\\x=-\dfrac{7}{2}+1\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=\dfrac{7}{2}+\dfrac{2}{2}\\x=-\dfrac{7}{2}+\dfrac{2}{2}\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=\dfrac{7+2}{2}\\x=\dfrac{-7+2}{2}\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=\dfrac{9}{2}\\x=\dfrac{-5}{2}\end{array} \right.\)
