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МатематикаГДЗ по алгебре 9 класс Макарычев ФГОС Задание 325
Задание 325
\[\boxed{\text{326.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ x^{4} - 20x^{2} + 64 = 0\]
\[Пусть\ t = x^{2},\ \]
\[t^{2} - 20x + 64 = 0\]
\[D = 100 - 64 = 36\]
\[t_{1,2} = 10 \pm 6 = 16;4;\]
\[\Longrightarrow x^{4} - 20x^{2} + 64 =\]
\[= \left( x^{2} - 16 \right)\left( x^{2} - 4 \right) =\]
\[= (x - 4)(x + 4)(x - 2)(x + 2).\]
\[\textbf{б)}\ x^{4} - 17x^{2} + 16 = 0\]
\[Пусть\ t = x^{2},\]
\[t^{2} - 17t + 16 = 0,\ по\ теореме\ \]
\[Виета:\]
\[t_{1} = 16,\ \ t_{2} = 1,\]
\[\Longrightarrow x^{4} - 17x^{2} + 16 =\]
\[= \left( x^{2} - 16 \right)\left( x^{2} - 1 \right) =\]
\[= (x - 4)(x + 4)(x - 1)(x + 1).\]
\[\textbf{в)}\ x^{4} - 5x^{2} - 36 = 0\]
\[Пусть\ x^{2} = t,\]
\[t^{2} - 5t - 36 = 0\]
\[D = 25 + 4 \cdot 36 = 169\]
\[t_{1,2} = \frac{5 \pm 13}{2}\]
\[t_{1} = 9,\ \ t_{2} = - 4,\]
\[\Longrightarrow x^{4} - 5x^{2} - 36 =\]
\[= \left( x^{2} + 4 \right)(x - 3)(x + 3).\]
\[\textbf{г)}\ x^{4} - 3x^{2} - 4 = 0\]
\[Пусть\ t = x^{2},\ \ t^{2} = x^{4},\]
\[t^{2} - 3t - 4 = 0,\ \ по\ теореме\ \]
\[Виета:\]
\[t_{1} = 4,\ \ t_{2} = - 1,\]
\[\Longrightarrow x^{4} - 3x^{2} - 4 =\]
\[= \left( x^{2} - 4 \right)\left( x^{2} + 1 \right) =\]
\[= (x - 2)(x + 2)\left( x^{2} + 1 \right).\]
\[\textbf{д)}\ 9x^{4} - 10x^{2} + 1 = 0\]
\[Пусть\ t = x^{2},\ \ t^{2} = x^{4},\]
\[9t^{2} - 10t + 1 = 0\]
\[D = 25 - 9 = 16\]
\[t = \frac{5 \pm 4}{9} = \frac{1}{9};1;\]
\[\Longrightarrow 9x^{4} - 10x^{2} + 1 =\]
\[= 9 \cdot \left( x^{2} - 1 \right)\left( x^{2} - \frac{1}{9} \right) =\]
\[= (x - 1)(x + 1)(3x - 1)(3x + 1).\ \]
\[\textbf{е)}\ \ 4x^{4} - 17x^{2} + 4 = 0\]
\[Пусть\ \ t = x^{2},\ \ t^{2} = x^{4},\]
\[4t^{2} - 17t + 4 = 0\]
\[D = 17^{2} - 4 \cdot 4 \cdot 4 = 225\]
\[t_{1,2} = \frac{17 \pm 15}{8} = \frac{1}{4};4;\ \]
\[\Longrightarrow 4x^{4} - 17x^{2} + 4 =\]
\[= 4 \cdot \left( x^{2} - 4 \right)\left( x^{2} - \frac{1}{4} \right) =\]
\[= \left( x^{2} - 4 \right)\left( 4x^{2} - 1 \right) =\]
\[= (x - 2)(x + 2)(2x - 1)(2x + 1).\]
