Đáp án: x=1;x=-1
Giải thích:
$\begin{array}{l}
{x^2} - {\left\{ {{{\left[ {{6^2} - {{\left( {{8^2} - 9.7} \right)}^3} - 7.5} \right]}^{ - 5}}.3} \right\}^3} = 1\\
\Leftrightarrow {x^2} - {\left\{ {{{\left[ {36 - {{\left( {64 - 63} \right)}^3} - 35} \right]}^{ - 5}}.3} \right\}^3} = 1\\
\Leftrightarrow {x^2} - {\left[ {{{\left( {1 - {1^3}} \right)}^{ - 5}}.3} \right]^3} = 1\\
\Leftrightarrow {x^2} - {\left( {{0^{ - 5}}.3} \right)^3} = 1\\
\Leftrightarrow {x^2} - {0^3} = 1\\
\Leftrightarrow {x^2} = 1\\
\Leftrightarrow x = 1;x = - 1\\
Vậy\,x = 1;x = - 1
\end{array}$
