Đáp án:
$\begin{array}{l}
d)\dfrac{{{2^{12}}{{.3}^5} - {4^6}.81}}{{{{\left( {{2^2}.3} \right)}^6} + {8^4}{{.3}^5}}}\\
= \dfrac{{{2^{12}}{{.3}^5} - {2^{12}}{{.3}^4}}}{{{2^{12}}{{.3}^6} + {2^{12}}{{.3}^5}}}\\
= \dfrac{{{2^{12}}{{.3}^4}\left( {3 - 1} \right)}}{{{2^{12}}{{.3}^5}.\left( {3 + 1} \right)}}\\
= \dfrac{2}{{3.4}}\\
= \dfrac{1}{6}\\
e)4.{\left( { - \dfrac{1}{2}} \right)^3} - 2.{\left( { - \dfrac{1}{2}} \right)^2} + 3.\left( { - \dfrac{1}{2}} \right) + 1\\
= {2^2}.\dfrac{{ - 1}}{{{2^3}}} - 2.\dfrac{1}{{{2^2}}} - \dfrac{3}{2} + 1\\
= \dfrac{{ - 1}}{2} - \dfrac{1}{2} - \dfrac{3}{2} + 1\\
= - 1 - \dfrac{3}{2} + 1\\
= - \dfrac{3}{2}\\
2)h){\left( {x - 1} \right)^2} = {\left( {x - 1} \right)^4}\\
\Rightarrow {\left( {x - 1} \right)^2}.\left( {1 - {{\left( {x - 1} \right)}^2}} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 1\\
{\left( {x - 1} \right)^2} = 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 1\\
x - 1 = 1\\
x - 1 = - 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 1\\
x = 2\\
x = 0
\end{array} \right.\\
Vậy\,x = 0;x = 1;x = 2\\
i){5^{ - 1}}{.25^x} = 125\\
\Rightarrow {5^{ - 1 + 2x}} = {5^3}\\
\Rightarrow - 1 + 2x = 3\\
\Rightarrow 2x = 4\\
\Rightarrow x = 2\\
Vậy\,x = 2\\
k)\left| {x + 1} \right| + \left| {x + 2} \right| + \left| {x + 3} \right| = 4x\\
+ Khi;x \ge - 1\\
\Rightarrow x + 1 + x + 2 + x + 3 = 4x\\
\Rightarrow x = 6\left( {tm} \right)\\
+ Khi: - 2 \le x < 1\\
\Rightarrow - x - 1 + x + 2 + x + 3 = 4x\\
\Rightarrow 3x = 6\\
\Rightarrow x = 2\left( {ktm} \right)\\
+ Khi: - 3 \le x < - 2\\
\Rightarrow - x - 1 - x - 2 + x + 3 = 4x\\
\Rightarrow 5x = 0\\
\Rightarrow x = 0\left( {ktm} \right)\\
+ khi:x < - 3\\
\Rightarrow - x - 1 - x - 2 - x - 3 = 4x\\
\Rightarrow 7x = - 6\\
\Rightarrow x = - \dfrac{6}{7}\left( {ktm} \right)\\
Vậy\,x = 6
\end{array}$
