вправа 521 гдз 7 клас алгебра Мерзляк Полонський
7 клас ➠ алгебра ➠ Мерзляк Полонський
Вправа 521
Відповідь:
\begin{equation}1)3^{20}\cdot 6^{20}-( 18^{10}-\end{equation}
\begin{equation}-2 )( 18^{10}+2 )=\end{equation}
\begin{equation}=18^{20}-\left ( 18^{20}-4 \right )=4;\end{equation}
\begin{equation}2)\left ( 5+28^{17} \right )\left ( 5-28^{17} \right )+\end{equation}
\begin{equation}+14^{34}\cdot 2^{34}=\end{equation}
\begin{equation}=25-28^{34}+28^{34}=25;\end{equation}
\begin{equation}3)7^{36}\cdot 8^{12}-\left ( 14^{18}+3 \right )\cdot \end{equation}
\begin{equation}\cdot \left ( 14^{18}-3 \right )=\end{equation}
\begin{equation}=7^{36}\cdot 2^{36}-\left ( 14^{36}-9 \right )=\end{equation}
\begin{equation}=14^{36}-14^{36}+9=9;\end{equation}
\begin{equation}4)\left ( 3^{2}-1 \right )\left ( 3^{2}+1 \right )\left ( 3^{4}+1 \right )\cdot \end{equation}
\begin{equation}\cdot \left ( 3^{8}+1 \right )\end{equation}
\begin{equation}\left ( 3^{16}+1 \right )\left ( 3^{32}+1 \right )-3^{64}=\end{equation}
\begin{equation}=\left ( 3^{4}-1 \right )\left ( 3^{4}+1 \right )\left ( 3^{8} +1\right )\end{equation}
\begin{equation}\left ( 3^{16}+1 \right )\left ( 3^{32}+1 \right )-3^{64}=\end{equation}
\begin{equation}=\left ( 3^{8}-1 \right )\left ( 3^{8}+1 \right )\left ( 3^{16} +1\right )\cdot \end{equation}
\begin{equation}\cdot \left ( 3^{32} +1\right )-3^{64}=\end{equation}
\begin{equation}=\left ( 3^{16}-1 \right )\left ( 3^{16}+1 \right )\cdot \end{equation}
\begin{equation}\cdot \left ( 3^{32}+1 \right )-3^{64}=\end{equation}
\begin{equation}=\left ( 3^{32}-1 \right )\left ( 3^{32}+1 \right )-3^{64}=\end{equation}
\begin{equation}=3^{64}-1-3^{64}=-1;\end{equation}
\begin{equation}5)\left ( 2+1 \right )\left ( 2^{2}+1 \right )\left ( 2^{4}+1 \right )\end{equation}
\begin{equation}\left ( 2^{8}+1 \right )\left ( 2^{16}+1 \right )-2^{32}=\end{equation}
\begin{equation}=\left ( 2-1 \right )\left ( 2+1 \right )\left ( 2^{2}+1 \right )\end{equation}
\begin{equation}\left ( 2^{4}+1 \right )\end{equation}
\begin{equation}\left ( 2^{8}+1 \right )\left ( 2^{16}+1 \right )-2^{32}=\end{equation}
\begin{equation}=\left ( 2^{2}-1 \right )\left ( 2^{2}+1 \right )\left ( 2^{4} +1\right )\end{equation}
\begin{equation}\left ( 2^{8} +1\right )\left ( 2^{16}+1 \right )-2^{32}=\end{equation}
\begin{equation}=\left ( 2^{4}-1 \right )\left ( 2^{4} +1\right )\left ( 2^{8} +1\right )\end{equation}
\begin{equation}\left ( 2^{16}+1 \right )-2^{32}=\end{equation}
\begin{equation}=\left ( 2^{8}-1 \right )\left ( 2^{8}+1 \right )\left ( 2^{16}+1 \right )-\end{equation}
\begin{equation}-2^{32}=\end{equation}
\begin{equation}=\left ( 2^{16}-1 \right )\left ( 2^{16}+1 \right )-2^{32}=\end{equation}
\begin{equation}=2^{32}-1-2^{32}=-1.\end{equation}