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МатематикаГДЗ по алгебре 8 класс Макарычев Задание 545
Задание 545
\[\boxed{\text{545\ (545).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
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\[\textbf{а)}\ (x + 4)^{2} = 3x + 40\] \[x^{2} + 8x + 16 = 3x + 40\] \[x^{2} + 5x - 24 = 0\] \[D = 25 + 96 = 121\] \[x_{1,2} = \frac{- 5 \pm \sqrt{121}}{2} = \frac{- 5 \pm 11}{2}\] \[x_{1} = - 8;\ \ x_{2} = 3.\] |
\[\textbf{б)}\ (2x - 3)^{2} = 11x - 19\] \[4x^{2} - 12x + 9 = 11x - 19\] \[4x^{2} - 23x + 28 = 0\] \[D = 529 - 448 = 81\] \[x_{1,2} = \frac{23 \pm \sqrt{81}}{2 \cdot 4} = \frac{23 \pm 9}{8}\] \[x_{1} = 4;\ \ x_{2} = 1,75.\] |
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\[\textbf{в)}\ 3(x + 4)^{2} = 10x + 32\] \[3\left( x^{2} + 8x + 16 \right) = 10x + 32\] \[3x^{2} + 24x + 48 - 10x - 32 = 0\] \[3x^{2} + 14x + 16 = 0\] \[D_{1} = 7^{2} - 3 \cdot 16 = 49 - 48 = 1\] \[x_{1,2} = \frac{- 7 \pm \sqrt{1}}{3} = \frac{- 7 \pm 1}{3}\] \[x_{1} = - \frac{8}{3} = - 2\frac{2}{3};\ \ x_{2} = - 2.\] |
\[\textbf{г)}\ 15x^{2} + 17 = 15(x + 1)^{2}\] \[15x^{2} + 17 = 15\left( x^{2} + 2x + 1 \right)\] \[15x^{2} + 17 - 15x^{2} - 30x - 15 = 0\] \[- 30x + 2 = 0\] \[- 30x = - 2\] \[x = \frac{1}{15}.\] |
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\[\textbf{д)}\ (x + 1)^{2} = 7918 - 2x\] \[x^{2} + 2x + 1 - 7918 + 2x = 0\] \[x^{2} + 4x - 7917 = 0\] \[D_{1} = 4 + 7917 = 7921\] \[x_{1,2} = - 2 \pm \sqrt{7921} = - 2 \pm 89\] \[x_{1} = 87;\ \ x_{2} = - 91.\] |
\[\textbf{е)}\ (x + 2)^{2} = 3131 - 2x\] \[x^{2} + 4x + 4 - 3131 + 2x = 0\] \[x^{2} + 6x - 3127 = 0\] \[D_{1} = 6^{2} + 3127 = 9 + 3127 = 3136\] \[x_{1,2} = - 3 \pm \sqrt{3136} = - 3 \pm 56\] \[x_{1} = - 59;\ \ x_{2} = 53.\] |
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\[\textbf{ж)}\ (x + 1)^{2} = (2x - 1)^{2}\] \[x^{2} - 2x + 1 = 4x^{2} - 4x + 1\] \[x^{2} - 2x + 1 - 4x^{2} + 4x - 1 = 0\] \[- 3x^{2} + 6x = 0\] \[- 3x(x - 2) = 0\] \[- 3x = 0\ \ \ \ \ \ \ x - 2 = 0\] \[\ \ \ \ \ \ x = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 2.\] |
\[\textbf{з)}\ (x - 2)^{2} + 48 = (2 - 3x)^{2}\] \[x^{2} - 4x + 4 + 48 = 4 - 12x + 9x^{2}\] \[x^{2} - 4x + 52 - 4 + 12x - 9x^{2} = 0\] \[- 8x^{2} + 8x + 48 = 0\ \ \ \ \ |\ :( - 8)\] \[x^{2} - x - 6 = 0\] \[D = 1 + 24 = 25\] \[x_{1,2} = \frac{1 \pm \sqrt{25}}{2} = \frac{1 \pm 5}{2}\] \[x_{1} = - 2;\ \ x_{2} = 3.\] |
