Help me . I love you chu cà mo ????????????????

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
\dfrac{{32}}{{{2^n}}} = 4 \Leftrightarrow {2^n} = 32:4 \Leftrightarrow {2^n} = 8 \Leftrightarrow {2^n} = {2^3} \Leftrightarrow n = 3\\
b,\\
\dfrac{{625}}{{{5^n}}} = 5 \Leftrightarrow {5^n} = 625:5 \Leftrightarrow {5^n} = 125 \Leftrightarrow {5^n} = {5^3} \Leftrightarrow n = 3\\
c,\\
{27^n}:{3^n} = 1\\
 \Leftrightarrow {\left( {{3^3}} \right)^n}:{3^n} = 1\\
 \Leftrightarrow {3^{3n}}:{3^n} = 1\\
 \Leftrightarrow {3^{3n – n}} = 1\\
 \Leftrightarrow {3^{2n}} = {3^0}\\
 \Leftrightarrow 2n = 0\\
 \Leftrightarrow n = 0\\
4,\\
\dfrac{{{4^6}{{.9}^5} + {6^9}.120}}{{{8^4}{{.3}^{12}} – {6^{11}}}}\\
 = \dfrac{{{{\left( {{2^2}} \right)}^5}.{{\left( {{3^2}} \right)}^5} + {{\left( {2.3} \right)}^9}{{.2}^3}.3.5}}{{{{\left( {{2^3}} \right)}^4}{{.3}^{12}} – {{\left( {2.3} \right)}^{11}}}}\\
 = \dfrac{{{2^{10}}{{.3}^{10}} + {2^9}{{.3}^9}{{.2}^3}.3.5}}{{{2^{12}}{{.3}^{12}} – {2^{11}}{{.3}^{11}}}}\\
 = \dfrac{{{2^{10}}{{.3}^{10}} + {2^{12}}{{.3}^{10}}.5}}{{{2^{11}}{{.3}^{11}}.\left( {2.3 – 1} \right)}}\\
 = \dfrac{{{2^{10}}{{.3}^{10}}.\left( {1 + {2^2}.5} \right)}}{{{2^{11}}{{.3}^{11}}.5}}\\
 = \dfrac{{{2^{10}}{{.3}^{10}}.21}}{{{2^{11}}{{.3}^{11}}.5}}\\
 = \dfrac{{21}}{{2.3.5}}\\
 = \dfrac{{21}}{{30}} = \dfrac{7}{{10}}\\
5,\\
a,\\
{x^2} – 0,25 = 0\\
 \Leftrightarrow {x^2} = 0,25\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0,5\\
x =  – 0,5
\end{array} \right.\\
b,\\
{x^3} + 27 = 0\\
 \Leftrightarrow {x^3} =  – 27\\
 \Leftrightarrow x =  – 3\\
c,\\
{\left( {\dfrac{1}{2}} \right)^x} = 64\\
 \Leftrightarrow \dfrac{1}{{{2^x}}} = {2^6}\\
 \Leftrightarrow {2^x}{.2^6} = 1\\
 \Leftrightarrow {2^{x + 6}} = {2^0}\\
 \Leftrightarrow x + 6 = 0\\
 \Leftrightarrow x =  – 6
\end{array}\)