a) `(3/8-1/5)+(5/8-x)=1/5`
⇔`7/(40)+5/8-x=1/5`
⇔`4/5-x=1/5`
⇔`x=3/5`
Vậy `x=3/5`
b) `2/(3).(x-3/4)+2.(x/3+2)=(23)/3`
⇔`2/3x-1/2+2/3x+4=(23)/3`
⇔`4/3x+7/2=(23)/3`
⇔`4/3x=(25)/6`
⇔`x=(25)/8`
Vậy `x=(25)/8`
c) `|x+3/4|+1/3=5/6`
⇔`|x+3/4|=1/2`
⇔\(\left[ \begin{array}{l}x+\dfrac{3}{4}=\dfrac{1}{2}\\x+\dfrac{3}{4}=-\dfrac{1}{2}\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-\dfrac{1}{4}\\x=-\dfrac{5}{4}\end{array} \right.\)
Vậy `x=-1/4` hoặc `x=-5/4`
d) `(1-x)/(-2)=(-8)/(1-x)`
⇔`(1-x)(1-x)=(-2).(-8)`
⇔`1-x-x+x^2=16`
⇔`1-2x+x^2=16`
⇔`(1-x)^2-16=0`
⇔(1-x-4)(1-x+4)=0`
⇔`(-3-x)(5-x)=0`
⇔\(\left[ \begin{array}{l}-3-x=0\\5-x=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-3\\x=5\end{array} \right.\)
Vậy `x=-3` hoặc `x=5`
