вправа 582 гдз 7 клас алгебра Мерзляк Полонський
7 клас ➠ алгебра ➠ Мерзляк Полонський
Вправа 582
Відповідь:
\begin{equation}1)\left ( 0,02p^{3}k+20p^{2}k^{4} \right )^{2}=\end{equation}
\begin{equation}=0,0004p^{6}k^{2}+0,8p^{5}k^{5}+400p^{4}k^{8};\end{equation}
\begin{equation}2)\left ( 1\frac{1}{6}mn-\frac{4}{21}m^{2}n^{5} \right )^{2}=\end{equation}
\begin{equation}=\left ( \frac{7}{6}mn-\frac{4}{21}m^{2}n^{5} \right )^{2}=\end{equation}
\begin{equation}=\frac{49}{36}m^{2}n^{2}-\frac{4}{9}m^{3}n^{6}+\frac{16}{441}m^{4}n^{10}=\end{equation}
\begin{equation}=1\frac{13}{36}m^{2}n^{2}-\frac{4}{9}m^{3}n^{6}+\frac{16}{441}m^{4}n^{10};\end{equation}
\begin{equation}3)-15\left ( \frac{1}{3}a-\frac{1}{5}b \right )^{2}=\end{equation}
\begin{equation}=-15\left ( \frac{1}{9}a^{2}-\frac{2}{15}ab+\frac{1}{25}b^{2} \right )=\end{equation}
\begin{equation}=-\frac{5}{3}a^{2}+2ab-\frac{3}{5}b^{2}=\end{equation}
\begin{equation}=-1\frac{2}{3}a^{2}+2ab-\frac{3}{5}b^{2};\end{equation}
\begin{equation}4)7x\left ( x^{3}-2x \right )^{2}=\end{equation}
\begin{equation}=7x\left ( x^{6}-4x^{4}+4x^{2} \right )=\end{equation}
\begin{equation}=7x^{7}-28x^{5}+28x^{3};\end{equation}
5) (5y - 2)²(2y + 1) =
= (25y² - 20y + 4)(2y + 1) =
= 50y³ - 40y² + 8y + 25y² - 20y + 4 =
= 50y³ - 15y² - 12y + 4; \begin{equation}6)\left ( 10p-k \right )^{2}\left ( 10p+k \right )^{2}=\end{equation} \begin{equation}=\left ( \left ( 10p-k \right )\left ( 10p+k \right ) \right )^{2}=\end{equation} \begin{equation}=\left ( 100p^{2}-k^{2} \right )^{2}=\end{equation} \begin{equation}=10000p^{4}-200p^{2}k^{2}+k^{4}.\end{equation}
= (25y² - 20y + 4)(2y + 1) =
= 50y³ - 40y² + 8y + 25y² - 20y + 4 =
= 50y³ - 15y² - 12y + 4; \begin{equation}6)\left ( 10p-k \right )^{2}\left ( 10p+k \right )^{2}=\end{equation} \begin{equation}=\left ( \left ( 10p-k \right )\left ( 10p+k \right ) \right )^{2}=\end{equation} \begin{equation}=\left ( 100p^{2}-k^{2} \right )^{2}=\end{equation} \begin{equation}=10000p^{4}-200p^{2}k^{2}+k^{4}.\end{equation}