вправа 355 гдз 7 клас алгебра Мерзляк Полонський
7 клас ➠ алгебра ➠ Мерзляк Полонський
Вправа 355
Відповідь:
1) 3x(2x + 5) = 6x² + 15x;
2) 4x(x² - 8x - 2) =
= 4x³ - 32x² - 8x;
3) -2a(a² + a - 3) =
= -2a³ - 2a² + 6a; \begin{equation}4)5b^{2}\left ( 3b^{2}-7b+10 \right )=\end{equation} \begin{equation}=15b^{4}-35b^{3}+50b^{2};\end{equation} \begin{equation}5)mn\left ( m^{2}n-n^{3} \right )=\end{equation} \begin{equation}=m^{3}n^{2}-mn^{4};\end{equation} \begin{equation}6)2ab\left ( a^{3}-3a^{2}b+b^{2} \right )=\end{equation} \begin{equation}=2a^{4}b-6a^{3}b^{2}+2ab^{3};\end{equation} \begin{equation}7)\left ( 4y^{3}-6y+7 \right )\cdot \left ( -1,2y^{3} \right )=\end{equation} \begin{equation}=-4,8y^{6}+7,2y^{4}-8,4y^{3};\end{equation} \begin{equation}8)0,4x^{2}y ( 3xy^{2}-\end{equation} \begin{equation}-5xy+13x^{2}y^{2} )=\end{equation} \begin{equation}=1,2x^{3}y^{3}-2x^{3}y^{2}+\end{equation} \begin{equation}+5,2x^{4}y^{3};\end{equation} \begin{equation}9)\left ( 2,3a^{3}b-1,7b^{4}-3,5b \right )\cdot \end{equation} \begin{equation}\left ( -10a^{2}b \right )=\end{equation} \begin{equation}=-23a^{5}b^{2}+17a^{2}b^{5}+\end{equation} \begin{equation}+35a^{2}b^{2};\end{equation} \begin{equation}10)-4pk^{3} ( 3p^{2}k-p+\end{equation} \begin{equation}+4k-2)=\end{equation} \begin{equation}=-12p^{3}k^{4}+4p^{2}k^{3}-\end{equation} \begin{equation}-16pk^{4}+8pk^{3};\end{equation} \begin{equation}11)\frac{2}{3}mn^{2}\left ( 6m-1,8n+9 \right )=\end{equation} \begin{equation}=4m^{2}n^{2}-1,2mn^{3}+6mn^{2};\end{equation} \begin{equation}12)1\frac{1}{7}cd ( \frac{7}{8}c^{5}-\end{equation} \begin{equation}-\frac{7}{24}c^{2}d^{7}-\frac{1}{4}d^{10} )=\end{equation} \begin{equation}=\frac{8}{7}cd\cdot \frac{7}{8}c^{5}-\frac{8}{7}cd\cdot \frac{7}{24}c^{2}d^{7}-\end{equation} \begin{equation}-\frac{8}{7}cd\cdot \frac{1}{4}d^{10}=\end{equation} \begin{equation}=c^{6}d-\frac{1}{3}c^{3}d^{8}-\frac{2}{7}cd^{11}.\end{equation}
1) 3x(2x + 5) = 6x² + 15x;
2) 4x(x² - 8x - 2) =
= 4x³ - 32x² - 8x;
3) -2a(a² + a - 3) =
= -2a³ - 2a² + 6a; \begin{equation}4)5b^{2}\left ( 3b^{2}-7b+10 \right )=\end{equation} \begin{equation}=15b^{4}-35b^{3}+50b^{2};\end{equation} \begin{equation}5)mn\left ( m^{2}n-n^{3} \right )=\end{equation} \begin{equation}=m^{3}n^{2}-mn^{4};\end{equation} \begin{equation}6)2ab\left ( a^{3}-3a^{2}b+b^{2} \right )=\end{equation} \begin{equation}=2a^{4}b-6a^{3}b^{2}+2ab^{3};\end{equation} \begin{equation}7)\left ( 4y^{3}-6y+7 \right )\cdot \left ( -1,2y^{3} \right )=\end{equation} \begin{equation}=-4,8y^{6}+7,2y^{4}-8,4y^{3};\end{equation} \begin{equation}8)0,4x^{2}y ( 3xy^{2}-\end{equation} \begin{equation}-5xy+13x^{2}y^{2} )=\end{equation} \begin{equation}=1,2x^{3}y^{3}-2x^{3}y^{2}+\end{equation} \begin{equation}+5,2x^{4}y^{3};\end{equation} \begin{equation}9)\left ( 2,3a^{3}b-1,7b^{4}-3,5b \right )\cdot \end{equation} \begin{equation}\left ( -10a^{2}b \right )=\end{equation} \begin{equation}=-23a^{5}b^{2}+17a^{2}b^{5}+\end{equation} \begin{equation}+35a^{2}b^{2};\end{equation} \begin{equation}10)-4pk^{3} ( 3p^{2}k-p+\end{equation} \begin{equation}+4k-2)=\end{equation} \begin{equation}=-12p^{3}k^{4}+4p^{2}k^{3}-\end{equation} \begin{equation}-16pk^{4}+8pk^{3};\end{equation} \begin{equation}11)\frac{2}{3}mn^{2}\left ( 6m-1,8n+9 \right )=\end{equation} \begin{equation}=4m^{2}n^{2}-1,2mn^{3}+6mn^{2};\end{equation} \begin{equation}12)1\frac{1}{7}cd ( \frac{7}{8}c^{5}-\end{equation} \begin{equation}-\frac{7}{24}c^{2}d^{7}-\frac{1}{4}d^{10} )=\end{equation} \begin{equation}=\frac{8}{7}cd\cdot \frac{7}{8}c^{5}-\frac{8}{7}cd\cdot \frac{7}{24}c^{2}d^{7}-\end{equation} \begin{equation}-\frac{8}{7}cd\cdot \frac{1}{4}d^{10}=\end{equation} \begin{equation}=c^{6}d-\frac{1}{3}c^{3}d^{8}-\frac{2}{7}cd^{11}.\end{equation}