cứuuuuuuuuuuuu $1)2x^2+(-6)^3:27=0$ $2)(x+\frac12)^2=\frac1{16}$ $3)(x-\frac13)^\prime-\frac{-8}{27}$,$4)(x-\frac29)^3=(\frac{-2}3)^6$ $5)(-\frac32)^x-\frac94$ $6)(\frac35)^n+\frac{98}{125}=1$,$7)(-\frac13)^{x-3}=\frac1{81}$ $8)5^{x+1}-5^x=20$ $9)4^{2x+1}+4^{2x}=80$

Question

Accepted Answer

$\displaystyle \begin{array}{{>{\displaystyle}l}}
1) \ 2x^{2} +( -6)^{3} :27=0\
\Longrightarrow 2x^{2} -8=0\
\Longrightarrow 2x^{2} =8\
\Longrightarrow x^{2} =4\
\Longrightarrow \left[ \begin{array}{l l}
x=2 & \
x=-2 &
\end{array} ight.\
2) \ \left( x+\frac{1}{2} ight)^{2} =\frac{1}{16}\
\Longrightarrow \left[ \begin{array}{l l}
x+\frac{1}{2} =\frac{1}{4} & \
x+\frac{1}{2} =\frac{-1}{4} &
\end{array} ight.\
\Longrightarrow \left[ \begin{array}{l l}
x=\frac{-1}{4} & \
x=\frac{-3}{4} &
\end{array} ight.\
3) \ \left( x-\frac{1}{3} ight)^{3} =\frac{-8}{27}\
\Longrightarrow x-\frac{1}{3} =\frac{-2}{3}\
\Longrightarrow x=\frac{-1}{3}\
4) \ \left( x-\frac{2}{9} ight)^{3} =\left(\frac{-2}{3} ight)^{6}\
\Longrightarrow \left( x-\frac{2}{9} ight)^{3} =\left(\frac{4}{9} ight)^{3}\
\Longrightarrow x-\frac{2}{9} =\frac{4}{9}\
\Longrightarrow x=\frac{2}{3}\
5) \ \left(\frac{-3}{2} ight)^{x} =\frac{9}{4}\
\Longrightarrow \left(\frac{-3}{2} ight)^{x} =\left( -\frac{3}{2} ight)^{2}\
\Longrightarrow x=2\
\end{array}$