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МатематикаРешебник по алгебре 9 класс Макарычев Задание 363
Задание 363
\[\boxed{\text{363\ (363).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ x^{4} - 20x^{2} + 64 = 0\]
\[Пусть\ t = x^{2}:\ \]
\[t^{2} - 20x + 64 = 0\]
\[D = 100 - 64 = 36\]
\[t_{1,2} = 10 \pm 6 = 16;4.\]
\[\Longrightarrow x^{4} - 20x^{2} + 64 =\]
\[= \left( x^{2} - 16 \right)\left( x^{2} - 4 \right) =\]
\[= (x - 4)(x + 4)(x - 2)(x + 2).\]
\[\textbf{б)}\ x^{4} - 17x^{2} + 16 = 0\]
\[Пусть\ t = x^{2}:\]
\[t^{2} - 17t + 16 = 0\]
\[По\ теореме\ Виета:\]
\[t_{1} + t_{2} = 17;\ \ \ t_{1} \cdot t_{2} = 16\]
\[t_{1} = 16;\ \ \ \ t_{2} = 1.\]
\[x^{4} - 17x^{2} + 16 = \left( x^{2} - 16 \right) \cdot\]
\[\cdot \left( x^{2} - 1 \right) =\]
\[= (x - 4)(x + 4)(x - 1)(x + 1).\]
\[\textbf{в)}\ x^{4} - 5x^{2} - 36 = 0\]
\[Пусть\ x^{2} = t:\]
\[t^{2} - 5t - 36 = 0\]
\[D = 25 + 4 \cdot 36 = 169\]
\[t_{1,2} = \frac{5 \pm 13}{2}\]
\[t_{1} = 9;\ \ \ t_{2} = - 4.\]
\[\Longrightarrow x^{4} - 5x^{2} - 36 =\]
\[= \left( x^{2} - 9 \right)\left( x^{2} + 4 \right) =\]
\[= \left( x^{2} + 4 \right)(x - 3)(x + 3).\]
\[\textbf{г)}\ x^{4} - 3x^{2} - 4 = 0\]
\[Пусть\ t = x^{2}:\]
\[t^{2} - 3t - 4 = 0\]
\[По\ теореме\ Виета:\]
\[t_{1} + t_{2} = 3;\ \ \ t_{1} \cdot t_{2} = - 4\]
\[t_{1} = 4;\ \ t_{2} = - 1.\]
\[\Longrightarrow x^{4} - 3x^{2} - 4 = \left( x^{2} - 4 \right) \cdot\]
\[\cdot \left( x^{2} + 1 \right) =\]
\[= (x - 2)(x + 2)\left( x^{2} + 1 \right).\]
\[\textbf{д)}\ 9x^{4} - 10x^{2} + 1 = 0\]
\[Пусть\ t = x^{2}:\]
\[9t^{2} - 10t + 1 = 0\]
\[D = 25 - 9 = 16\]
\[t = \frac{5 \pm 4}{9} = \frac{1}{9};1.\]
\[\Longrightarrow 9x^{4} - 10x^{2} + 1 =\]
\[= 9 \cdot \left( x^{2} - 1 \right)\left( x^{2} - \frac{1}{9} \right) =\]
\[= (x - 1)(x + 1)(3x - 1) \cdot\]
\[\cdot (3x + 1).\ \]
\[\textbf{е)}\ \ 4x^{4} - 17x^{2} + 4 = 0\]
\[Пусть\ \ t = x^{2}:\]
\[4t^{2} - 17t + 4 = 0\]
\[D = 17^{2} - 4 \cdot 4 \cdot 4 = 225\]
\[t_{1,2} = \frac{17 \pm 15}{8} = \frac{1}{4};4.\ \]
\[\Longrightarrow 4x^{4} - 17x^{2} + 4 = 4 \cdot \left( x^{2} - 4 \right) \cdot\]
\[\cdot \left( x^{2} - \frac{1}{4} \right) = \left( x^{2} - 4 \right)\left( 4x^{2} - 1 \right) =\]
\[= (x - 2)(x + 2)(2x - 1)(2x + 1).\]
\[\boxed{\text{363.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[(y - 5x)(x + y + 4) = 0\]
\[1)\ y - 5x = 0\]
\[y = 5x.\]
\[2)\ x + y + 4 = 0\]
\[y = - x - 4.\]
