Bài 6.
a)
`|x-1,7|=2,5`
⇔\(\left[ \begin{array}{l}x-1,7=2,5\\x-1,7=-2,5\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=2,5+1,7\\x=-2,5+1,7\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=4,2\\x=-0,8\end{array} \right.\)
Vậy `x=4,2` hoặc `x=-0,8`
b)
`|x+\frac{3}{4}|-\frac{6}{5}=0`
`⇔|x+\frac{3}{4}|=\frac{6}{5}`
⇔\(\left[ \begin{array}{l}x+\dfrac{3}{4}=\dfrac{6}{5}\\x+\dfrac{3}{4}=-\dfrac{6}{5}\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\dfrac{6}{5}-\dfrac{3}{4}\\x=-\dfrac{6}{5}-\dfrac{3}{4}\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\dfrac{9}{20}\\x=-\dfrac{39}{20}\end{array} \right.\)
Vậy $x=\dfrac{9}{20},x=-\dfrac{39}{20}$
c)
`|x|=1\frac{2}{5}`
`⇔|x|=\frac{7}{5}`
⇔\(\left[ \begin{array}{l}x=\dfrac{7}{5}\\x=-\dfrac{7}{5}\end{array} \right.\)
Vậy $x=\dfrac{7}{5},x=-\dfrac{7}{5}$
