вправа 376 гдз 7 клас алгебра Мерзляк Полонський
7 клас ➠ алгебра ➠ Мерзляк Полонський
Вправа 376
Відповідь:
\begin{equation}1)x-\frac{7x+1}{8}=\frac{4x+3}{4};\end{equation}
\begin{equation}\frac{8x}{8}-\frac{7x+1}{8}=\frac{2\cdot \left ( 4x+3 \right )}{8};\end{equation}
8x - 7x - 1 = 8x + 6;
-7x = 7;
x = -1; \begin{equation}2)\frac{2x+1}{6}-\frac{3x+1}{7}=2;\end{equation} \begin{equation}\frac{7\cdot \left ( 2x+1 \right )}{42}-\frac{6\cdot \left ( 3x+1 \right )}{42}=\end{equation} \begin{equation}=\frac{42\cdot 2}{42};\end{equation} 14x + 7 - 18x - 6 = 84;
-4x = 83; \begin{equation}x=-20\frac{3}{4};\end{equation} \begin{equation}3)\frac{2x+3}{3}-\frac{5x+13}{6}+\end{equation} \begin{equation}+\frac{5-2x}{2}=6;\end{equation} \begin{equation}\frac{2\cdot \left ( 2x+3 \right )}{6}-\frac{5x+13}{6}+\end{equation} \begin{equation}+\frac{3\cdot \left ( 5-2x \right )}{6}=\end{equation} \begin{equation}=\frac{6\cdot 6}{6};\end{equation} 4x + 6 - 5x - 13 + 15 - 6x = 36;
-7x = 28;
x = -4; \begin{equation}4)\frac{4x^{2}+5x}{14}+\frac{10-2x^{2}}{7}=5;\end{equation} \begin{equation}\frac{4x^{2}+5x}{14}+\frac{2\cdot \left ( 10-2x^{2} \right )}{14}=\end{equation} \begin{equation}=\frac{5\cdot 14}{14};\end{equation} 4x² + 5x + 20 - 4x² = 70;
5x = 50;
x = 10.
-7x = 7;
x = -1; \begin{equation}2)\frac{2x+1}{6}-\frac{3x+1}{7}=2;\end{equation} \begin{equation}\frac{7\cdot \left ( 2x+1 \right )}{42}-\frac{6\cdot \left ( 3x+1 \right )}{42}=\end{equation} \begin{equation}=\frac{42\cdot 2}{42};\end{equation} 14x + 7 - 18x - 6 = 84;
-4x = 83; \begin{equation}x=-20\frac{3}{4};\end{equation} \begin{equation}3)\frac{2x+3}{3}-\frac{5x+13}{6}+\end{equation} \begin{equation}+\frac{5-2x}{2}=6;\end{equation} \begin{equation}\frac{2\cdot \left ( 2x+3 \right )}{6}-\frac{5x+13}{6}+\end{equation} \begin{equation}+\frac{3\cdot \left ( 5-2x \right )}{6}=\end{equation} \begin{equation}=\frac{6\cdot 6}{6};\end{equation} 4x + 6 - 5x - 13 + 15 - 6x = 36;
-7x = 28;
x = -4; \begin{equation}4)\frac{4x^{2}+5x}{14}+\frac{10-2x^{2}}{7}=5;\end{equation} \begin{equation}\frac{4x^{2}+5x}{14}+\frac{2\cdot \left ( 10-2x^{2} \right )}{14}=\end{equation} \begin{equation}=\frac{5\cdot 14}{14};\end{equation} 4x² + 5x + 20 - 4x² = 70;
5x = 50;
x = 10.