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МатематикаГДЗ по алгебре 8 класс Макарычев Задание 600
Задание 600
\[\boxed{\text{600\ (600).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
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\[\textbf{а)}\ \frac{y^{2}}{y + 3} = \frac{y}{y + 3}\] \[y^{2} = y\ при\ y + 3 \neq 0\] \[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y \neq - 3\] \[y^{2} - y = 0\] \[y(y - 1) = 0\] \[y = 0\ \ \ \ \ \ y - 1 = 0\] \[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y = 1\] \[Ответ:y = \left\{ 0;1 \right\}.\] |
\[\textbf{б)}\ \frac{x^{2}}{x^{2} - 4} = \frac{5x - 6}{x^{2} - 4}\] \[x^{2} - 5x - 6\ \ при\ x^{2} - 4 \neq 0\] \[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }(x - 2)(x + 2) \neq 0\] \[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \neq \pm 2\] \[x^{2} - 5x + 6 = 0\] \[x_{1} + x_{2} = 5;\ \ \ x_{1} \cdot x_{2} = 6\] \[x = 3;\ \ \ \ \ x = 2.\ \ \] \[так\ как\ x \neq 2\] \[Ответ:x = 3.\] |
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\[\textbf{в)}\frac{2x^{2}}{x - 2} = \frac{- 7x + 6}{2 - x}\] \[\frac{2x^{2}}{x - 2} = \frac{- (7x - 6)}{- (x - 2)}\] \[2x^{2} = 7x - 6\ \ \ \ \ при\ x - 2 \neq 0\] \[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \neq 2\] \[2x^{2} - 7x + 6 = 0\] \[D = 49 - 48 = 1\] \[x_{1,2} = \frac{7 \pm 1}{4} = 2;1,5\] \[x = 1,5;\] \[так\ как\ x \neq 2\] \[Ответ:x = 1,5.\] |
\[\textbf{г)}\ \frac{y^{2} - 6y}{y - 5} = \frac{5}{5 - y}\] \[\frac{y^{2} - 6y}{y - 5} = \frac{- 5}{y - 5}\] \[y^{2} - 6y = - 5\ \ \ \ при\ y - 5 \neq 0\] \[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y \neq 5\] \[y^{2} - 6y + 5 = 0\] \[y_{1} + y_{2} = 6;\ \ \ y_{1} \cdot y_{2} = 5\] \[y = 1;\ \ \ \ \ \ \ y = 5\] \[так\ как\ y \neq 5\] \[Ответ:y = 1.\] |
\[\textbf{д)}\frac{2x - 1}{x + 7} = \frac{3x + 4}{x - 1}\text{\ \ \ \ \ }\]
\[при\ x + 7 \neq 0\ \ \ \ \ и\ \ \ \ \ x - 1 \neq 0\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \neq - 7\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \neq 1\]
\[(2x - 1)(x - 1) =\]
\[= (3x + 4)(x + 7)\]
\[2x^{2} - 2x - x + 1 =\]
\[= 3x^{2} + 21x + 4x + 28\]
\[x^{2} + 28x + 27 = 0\]
\[D = 784 - 108 = 676 = 26^{2}\]
\[x_{1,2} = \frac{- 28 \pm 26}{2} = - 1;\ - 27\]
\[Ответ:x = \left\{ - 27;\ - 1 \right\}.\]
\[\textbf{е)}\frac{2y + 3}{2y - 1} = \frac{y - 5}{y + 3}\]
\[при\ 2y - 1 \neq 0\ \ \ \ и\ \ \ \ y + 3 \neq 0\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y \neq \frac{1}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y \neq - 3\]
\[(2y + 3)(y + 3) =\]
\[= (y - 5)(2y - 1)\]
\[2y^{2} + 6y + 3y + 9 =\]
\[= 2y^{2} - y - 10y + 5\]
\[20y + 4 = 0\]
\[20y = - 4\]
\[y = - \frac{4}{20}\]
\[y = - \frac{1}{5}\]
\[Ответ:y = - 0,2.\]
\[\textbf{ж)}\frac{5y + 1}{y + 1} = \frac{y + 2}{y}\]
\[при\ y + 1 \neq 0\ \ и\ \ \ \ \ y \neq 0\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y \neq - 1\]
\[(5y + 1)y = (y + 2)(y + 1)\]
\[5y^{2} + y = y^{2} + y + 2y + 2\]
\[4y^{2} - 2y - 2 = 0\ \ \ \ \ \ |\ :2\]
\[2y^{2} - y - 1 = 0\]
\[D = 1 + 8 = 9\]
\[y_{1,2} = \frac{1 \pm 3}{4} = 1;\ - \frac{1}{2}\]
\[Ответ:y = \left\{ - 0,5;1 \right\}.\]
\[\textbf{з)}\frac{1 + 3x}{1 - 2x} = \frac{5 - 3x}{1 + 2x}\]
\[при\ 1 - 2x \neq 0\ \ \ и\ \ \ \ 1 + 2x \neq 0\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \neq \frac{1}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \neq - \frac{1}{2}\]
\[(1 + 3x)(1 + 2x) =\]
\[= (5 - 3x)(1 - 2x)\]
\[1 + 2x + 3x + 6x^{2} =\]
\[= 5 - 10x - 3x + 6x^{2}\]
\[18x - 4 = 0\]
\[18x = 4\]
\[x = \frac{4}{18}\]
\[x = \frac{2}{9}\]
\[Ответ:x = \frac{2}{9}.\]
\[\textbf{и)}\frac{x - 1}{2x + 3} - \frac{2x - 1}{3 - 2x} = 0\]
\[\frac{x - 1}{2x + 3} = \frac{2x - 1}{3 - 2x}\]
\[при\ 2x + 3 \neq 0\ \ \ и\ \ \ 3 - 2x \neq 0\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2x \neq - 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ 2x \neq 3\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \neq - 1,5\ \ \ \ \ \ \ \ \ \ \ \ \ x \neq 1,5\]
\[(x - 1)(3 - 2x) =\]
\[= (2x - 1)(2x + 3)\]
\[3x - 2x^{2} - 3 + 2x =\]
\[= 4x^{2} + 6x - 2x - 3\]
\[6x^{2} - x = 0\]
\[x(6x - 1) = 0\]
\[x = 0,\ \ 6x - 1 = 0\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6x = 1\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = \frac{1}{6}\]
\[Ответ:x = \left\{ 0;\frac{1}{6} \right\}.\]
