Đáp án:
\(\begin{array}{l}
C5:\\
a)4{x^{16}}{y^{14}}{z^{12}}\\
b){a^{2nk}}{b^{2nk + k}}{c^{2nk + n}}\\
C6:\\
\left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
C5:\\
a){\left( {2{x^2}{y^3}{z^4}} \right)^2}.{\left( { - {x^3}{y^2}z} \right)^4}\\
= 4{x^4}{y^{3.2}}{z^{4.2}}.{x^{3.4}}.{y^{2.4}}.{z^4}\\
= 4{x^{4 + 12}}.{y^{6 + 8}}.{z^{8 + 4}}\\
= 4{x^{16}}{y^{14}}{z^{12}}\\
b){\left( {{a^n}{b^{n + 1}}{c^n}} \right)^k}.{\left( {{a^k}{b^k}{c^{k + 1}}} \right)^n}\\
= {a^{n.k}}.{b^{nk + k}}.{c^{nk}}.{a^{nk}}.{b^{nk}}.{c^{nk + n}}\\
= {a^{2nk}}{b^{nk + k + nk}}.{c^{nk + nk + n}}\\
= {a^{2nk}}{b^{2nk + k}}{c^{2nk + n}}\\
C6:\\
{x^n} - 2{x^{n + 1}} + 5{x^n} - 4{x^{n + 1}} = 0\\
\to {x^n} - 2{x^n}.x + 5{x^n} - 4{x^n}.x = 0\\
\to 6{x^n} - 6{x^n}.x = 0\\
\to 6{x^n}\left( {1 - x} \right) = 0\\
\to \left[ \begin{array}{l}
{x^n} = 0\\
1 - x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.
\end{array}\)
