вправа 1149 гдз 7 клас алгебра Мерзляк Полонський

Відповідь: \begin{equation}1)2,2b-0,75b^{2}+\left ( \frac{1}{4}b^{2}-1\frac{4}{5}b \right )=\end{equation} \begin{equation}=2\frac{1}{5}b-\frac{3}{4}b^{2}+\frac{1}{4}b^{2}-1\frac{4}{5}b=\end{equation} \begin{equation}=\frac{2}{5}b-\frac{1}{2}b^{2};\end{equation} \begin{equation}2,2b-0,75b^{2}-\left ( \frac{1}{4}b^{2}-1\frac{4}{5}b \right )=\end{equation} \begin{equation}=2\frac{1}{5}b-\frac{3}{4}b^{2}-\frac{1}{4}b^{2}+1\frac{4}{5}b=\end{equation} \begin{equation}=4b-b^{2};\end{equation} \begin{equation}2)1\frac{2}{7}x^{2}+2\frac{4}{9}y+\left ( 2\frac{3}{14}x^{2}-1\frac{1}{6}y \right )=\end{equation} \begin{equation}=1\frac{2}{7}x^{2}+2\frac{4}{9}y+2\frac{3}{14}x^{2}-1\frac{1}{6}y=\end{equation} \begin{equation}=1\frac{4}{14}x^{2}+2\frac{8}{18}y+2\frac{3}{14}x^{2}-1\frac{3}{18}y=\end{equation} \begin{equation}=3\frac{7}{14}x^{2}-1\frac{5}{18}y=\end{equation} \begin{equation}=3\frac{1}{2}x^{2}-1\frac{5}{18}y;\end{equation} \begin{equation}1\frac{2}{7}x^{2}+2\frac{4}{9}y-\left ( 2\frac{3}{14}x^{2}-1\frac{1}{6}y \right )=\end{equation} \begin{equation}=1\frac{2}{7}x^{2}+2\frac{4}{9}y-2\frac{3}{14}x^{2}+1\frac{1}{6}y=\end{equation} \begin{equation}=1\frac{4}{14}x^{2}+2\frac{8}{18}y-2\frac{3}{14}x^{2}+1\frac{3}{18}y=\end{equation} \begin{equation}=-\frac{13}{14}x^{2}+3\frac{11}{18}y.\end{equation}