Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{5^x}:5 = 125\\
\Leftrightarrow {5^{x - 1}} = {5^3}\\
\Leftrightarrow x - 1 = 3\\
\Leftrightarrow x = 3 + 1\\
\Leftrightarrow x = 4\\
b,\\
{9^2}.{\left( {5x - 2} \right)^3} = {3^7}\\
\Leftrightarrow {\left( {5x - 2} \right)^3} = {3^7}:{9^2}\\
\Leftrightarrow {\left( {5x - 2} \right)^3} = {3^7}:{\left( {{3^2}} \right)^2}\\
\Leftrightarrow {\left( {5x - 2} \right)^3} = {3^7}:{3^4}\\
\Leftrightarrow {\left( {5x - 2} \right)^3} = {3^{7 - 4}}\\
\Leftrightarrow {\left( {5x - 2} \right)^3} = {3^3}\\
\Leftrightarrow 5x - 2 = 3\\
\Leftrightarrow 5x = 3 + 2\\
\Leftrightarrow 5x = 5\\
\Leftrightarrow x = 5:5\\
\Leftrightarrow x = 1\\
d,\\
{x^{2008}} = {x^5}\\
\Leftrightarrow {x^{2008}} - {x^5} = 0\\
\Leftrightarrow {x^{5 + 2003}} - {x^5} = 0\\
\Leftrightarrow {x^5}.{x^{2003}} - {x^5} = 0\\
\Leftrightarrow {x^5}.\left( {{x^{2003}} - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
{x^5} = 0\\
{x^{2003}} - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^{2003}} = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\\
1,\\
a,\\
36:\left\{ {390:\left[ {500 - \left( {195 + 25.7} \right)} \right]} \right\}\\
= 36:\left\{ {390:\left[ {500 - \left( {195 + 175} \right)} \right]} \right\}\\
= 36:\left\{ {390:\left[ {500 - 370} \right]} \right\}\\
= 36:\left\{ {390:130} \right\}\\
= 36:3\\
= 12\\
b,\\
381 + 107 - \left( {18.204 + 53.18} \right):257\\
= 488 - \left[ {18.\left( {204 + 53} \right)} \right]:257\\
= 488 - \left[ {18.257} \right]:257\\
= 488 - 18\\
= 470\\
c,\\
80 - \left( {{{4.5}^2} - {{3.2}^3}} \right)\\
= 80 - \left( {4.25 - 3.8} \right)\\
= 80 - \left( {100 - 24} \right)\\
= 80 - 76\\
= 4\\
d,\\
150 - \left[ {120 - {{\left( {7 - 2} \right)}^2}} \right]\\
= 150 - \left( {120 - {5^2}} \right)\\
= 150 - \left( {120 - 25} \right)\\
= 150 - 95\\
= 55
\end{array}\)
