$\\$
Bài `5.`
`a,`
`|3 - 2x|=1/2`
`->` \(\left[ \begin{array}{l}3-2x=\dfrac{1}{2}\\3-2x=\dfrac{-1}{2}\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}2x=3-\dfrac{1}{2}\\2x=3+\dfrac{1}{2}\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}2x=\dfrac{5}{2}\\2x=\dfrac{7}{2}\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}x=\dfrac{5}{2} :2\\x=\dfrac{7}{2}:2\end{array} \right.\) $\\$ `->` \(\left[ \begin{array}{l}x=\dfrac{5}{4}\\x=\dfrac{7}{4}\end{array} \right.\)
Vậy `x=5/4` hoặc `x=7/4`
`b,`
`(x-1)/4=10/8`
`-> x-1=4 . 10/8`
`->x-1=5`
`->x=5+1`
`->x=6`
Vậy `x=6`
$\\$
Bài `6.`
Đặt `a/b=c/d=k (k \ne 0)`
`->` $\begin{cases} \dfrac{a}{b}=k\\\dfrac{c}{d}=k \end{cases}$ `->` $\begin{cases} a=bk\\c=dk \end{cases}$
Có : `(a+c)/(b+d)`
`= (bk + dk)/(b+d)`
`= (k (b+d) )/(b+d)`
`= k` `(1)`
Có : `(a-c)/(b-d)`
`= (bk - dk)/(b-d)`
`= (k (b-d) )/(b-d)`
`= k` `(2)`
Từ `(1), (2)`
`-> (a+c)/(b+d) = (a-c)/(b-d) (=k)`
`->` đpcm
