Bài 1:
a) $\frac{1}{4}$.x-$\frac{1}{3}$=-$\frac{5}{9}$
$\frac{1}{4}$.x =-$\frac{5}{9}$+$\frac{1}{3}$
$\frac{1}{4}$.x =$\frac{-5+3}{9}$
$\frac{1}{4}$.x =$\frac{-2}{9}$
x =$\frac{-2}{9}$÷$\frac{1}{4}$
x =$\frac{-2}{9}$.4
x =$\frac{-8}{9}$
b) $\frac{11}{12}$-($\frac{2}{5}$-x)=$\frac{3}{4 }$
$\frac{2}{5}$-x =$\frac{11}{12}$-$\frac{3}{4 }$
$\frac{2}{5}$-x =$\frac{11-9}{12}$
$\frac{2}{5}$-x =$\frac{-2}{9}$
$\frac{2}{5}$-x =$\frac{-1}{6}$
-x =$\frac{-1}{6}$-$\frac{2}{5}$
-x =$\frac{5-12}{30}$
-x =$\frac{-7}{30}$
x =$\frac{7}{30}$
d) lx+$\frac{1}{3}$l-4=-1
lx+$\frac{1}{3}$l =-1+4
lx+$\frac{1}{3}$l = 3
⇒$\left \{ {{x+\frac{1}{3}=3} \atop {x+\frac{1}{3}=-3}} \right.$
⇒$\left \{ {{x=3-\frac{1}{3}} \atop {x=-3-\frac{1}{3}}} \right.$
⇒$\left \{ {{x=\frac{9-1}{3} } \atop {x=\frac{-9-1}{3} }} \right.$
⇒$\left \{ {{x=\frac{8}{3}} \atop {x=\frac{-10}{3} }} \right.$