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номер 126 гдз 8 класс алгебра Муравин Муравин

1)\begin{equation}\frac{x-7}{3x+5}+\frac{2x+6}{3x-1}=1 \end{equation}
ОДЗ: \begin{equation}3x+5\neq 0 \end{equation} \begin{equation}3x\neq -5 \end{equation} \begin{equation}x\neq -\frac{5}{3} \end{equation} \begin{equation}3x-1\neq 0 \end{equation} \begin{equation}3x\neq 1 \end{equation} \begin{equation}x\neq \frac{1}{3} \end{equation} \begin{equation}\frac{x-7}{3x+5}+\frac{2x+6}{3x-1}-1=0 \end{equation} \begin{equation}\frac{\left ( 3x-1 \right )\left ( x-7 \right )+\left ( 3x+5 \right )\left ( 2x+6 \right )-\left ( 3x+5 \right )\left ( 3x-1 \right )}{\left ( 3x+5 \right )\left ( 3x-1 \right )}=0 \end{equation}
(3x - 1)(x - 7) + (3x + 5)(2x + 6) - (3x + 5)(3x - 1) = 0
3x² - 21x - x + 7 + 6x² + 18x + 10x + 30 - (9x² - 3x + 15x - 5) = 0
3x² - 22x + 7 + 6x² + 28x + 30 - 9x² - 12x + 5 = 0
-6x + 42 = 0
-6x = -42 \begin{equation}x=\frac{-42}{-6} \end{equation}
x = 7
Ответ: x = 7
2)\begin{equation}\frac{3y-4}{6y+5}-\frac{1-y}{2y-1}=1 \end{equation} \begin{equation}\frac{3y-4}{6y+5}-\frac{1-y}{2y-1}-1=0 \end{equation} \begin{equation}\frac{\left ( 2y-1 \right )\left ( 3y-4 \right )-\left ( 1-y \right )\left ( 6y+5 \right )-\left ( 6y+5 \right )\left ( 2y-1 \right )}{\left ( 6y+5 \right )\left ( 2y-1 \right )}=0 \end{equation}
ОДЗ: \begin{equation}\left ( 6y+5 \right )\left ( 2y-1 \right )\neq 0 \end{equation} \begin{equation}6y+5\neq 0 \end{equation} \begin{equation}6y\neq -5 \end{equation} \begin{equation}y\neq - \frac{5}{6} \end{equation} \begin{equation}2y-1\neq 0 \end{equation} \begin{equation}2y\neq1 \end{equation} \begin{equation}y\neq \frac{1}{2} \end{equation}
(2y - 1)(3y - 4) - (1 - y)(6y + 5) - (6y + 5)(2y - 1) = 0
6y² - 8y - 3y + 4 - (6y + 5 - 6y² - 5y) - (12y² - 6y + 10y - 5) = 0
6y² - 8y - 3y + 4 - 6y - 5 + 6y² + 5y - 12y² + 6y - 10y + 5 = 0
-16y + 4 = 0
-16y = -4 \begin{equation}y=\frac{-4}{-16} \end{equation} \begin{equation}y=\frac{1}{4} \end{equation}
3)\begin{equation}\frac{1}{y-3}\cdot \frac{4}{y+1}=\frac{1}{y-3}+\frac{4}{y+1} \end{equation}
ОДЗ: \begin{equation}y-3\neq 0 \end{equation} \begin{equation}y\neq 3 \end{equation} \begin{equation}y+1\neq 0 \end{equation}
y = -1 \begin{equation}\frac{4}{\left ( y-3 \right )\left ( y+1 \right )}=\frac{1}{y-3}+\frac{4}{y+1} \end{equation}
4 = y + 1 + 4(y - 3)
4 = y + 1 + 4y - 12
4 - 1 + 12 = y + 4y
15 = 5y
y = 3
Ответ: решений нет
4)\begin{equation}\frac{5p-2}{2p-1}\cdot \frac{5p+1}{7p-3}=\frac{5p-2}{2p-1}-\frac{5p+1}{7p-3} \end{equation}
ОДЗ: \begin{equation}2p-1\neq 0 \end{equation} \begin{equation}2p\neq 1 \end{equation} \begin{equation}p=\frac{1}{2} \end{equation} \begin{equation}7p-3\neq 0 \end{equation} \begin{equation}7p\neq 3 \end{equation} \begin{equation}p=\frac{3}{7} \end{equation} \begin{equation}\frac{5p-2}{2p-1}\cdot \frac{5p+1}{7p-3}-\frac{5p-2}{2p-1}+\frac{5p+1}{7p-3}=0 \end{equation}
(5p - 2)(5p + 1) - (7p - 3)(5p - 2) + (5p + 1)(2p - 1) = 0
25p² + 5p - 10p - 2 - (35p² - 14p - 15p + 6) + 10p² - 5p + 2p - 1 = 0
25p² - 5p - 2 - 35p² + 29p - 6 + 10p² - 3p - 1 = 0
21p - 9 = 0
21p = 9 \begin{equation}p=\frac{9}{21} \end{equation} \begin{equation}p=\frac{3}{7} \end{equation}
Ответ: решений нет