b, $\frac{11}{12}.x+$ $\frac{3}{4}=$ $\frac{1}{6}$ =>$\frac{11}{12}.x=$$\frac{-7}{12}$ =>$x=\frac{-14}{121}$
c, $2^x+4.2^x=5$
$=>2^x.(1+4)=5$
$=>2^x=1$
$=>2^x=2^0$
$=>x=0$
d, $(3x-4)(x-1)^3=0$
$=>$\(\left[ \begin{array}{l}3x-4=0\\x-1=0\end{array} \right.\) =>\(\left[ \begin{array}{l}3x=4\\x=1\end{array} \right.\) =>\(\left[ \begin{array}{l}x=4/3\\x=1\end{array} \right.\)
Vậy x=4/3 hoặc x=1
a, =>$x-\frac{7}{15}=$ $\frac{-21}{20}$ =>x=$\frac{-7}{12}$
