Tìm x biết a,32-/-2^n=4 b,(1/2)^2n-1=1/8 c,4/2=4/n/2 d,(n+5)^3=-64 Tính A=2^10+4^10/8^4+4^11 lm cho mik vs nek Thank you nhá vote luôn

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
32:{2^n} = 4\\
 \Leftrightarrow {2^n} = 32:4\\
 \Leftrightarrow {2^n} = 8\\
 \Leftrightarrow {2^n} = {2^3}\\
 \Leftrightarrow n = 3\\
b,\\
{\left( {\dfrac{1}{2}} \right)^{2n – 1}} = \dfrac{1}{8}\\
 \Leftrightarrow \dfrac{1}{{{2^{2n – 1}}}} = \dfrac{1}{8}\\
 \Leftrightarrow {2^{2n – 1}} = 8\\
 \Leftrightarrow {2^{2n – 1}} = {2^3}\\
 \Leftrightarrow 2n – 1 = 3\\
 \Leftrightarrow 2n = 4\\
 \Leftrightarrow n = 2\\
d,\\
{\left( {n + 5} \right)^3} =  – 64\\
 \Leftrightarrow {\left( {n + 5} \right)^3} = {\left( { – 4} \right)^3}\\
 \Leftrightarrow n + 5 =  – 4\\
 \Leftrightarrow n =  – 9\\
2,\\
A = \dfrac{{{2^{10}} + {4^{10}}}}{{{8^4} + {4^{11}}}} = \dfrac{{{2^{10}} + {{\left( {{2^2}} \right)}^{10}}}}{{{{\left( {{2^3}} \right)}^4} + {{\left( {{2^2}} \right)}^{11}}}} = \dfrac{{{2^{10}} + {2^{20}}}}{{{2^{12}} + {2^{22}}}} = \dfrac{{{2^{10}}.\left( {1 + {2^{10}}} \right)}}{{{2^{12}}.\left( {1 + {2^{10}}} \right)}} = \dfrac{1}{{{2^2}}} = \dfrac{1}{4}
\end{array}\)

\(\begin{array}{l}
c,\\
\dfrac{4}{2} = \dfrac{4}{{\dfrac{n}{2}}}\\
 \Leftrightarrow 2 = \dfrac{4}{{\dfrac{n}{2}}}\\
 \Leftrightarrow \dfrac{n}{2} = 4:2\\
 \Leftrightarrow \dfrac{n}{2} = 2\\
 \Leftrightarrow n = 4
\end{array}\)