Ответ:

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Объяснение:

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а)

[tex] {x}^{4} - 4 {x}^{2} + 3 = 0 \\ {x}^{2} = a \: , \: \: \: a \geqslant 0 \\ {a}^{2} - 4a + 3 = 0 \\ D = ( - 4 ){}^{2} - 4 \times 3 = 16 - 12 = 4 \\ a_{1} = \frac{4 - 2}{2} = \frac{2}{2} = 1\\ a _{2}= \frac{4 + 2}{2} = \frac{6}{2} = 3 \\ \\ {x}^{2} = 1 \\ {x}^{2} = 3 \\ \\ x_{1} = - 1 \\ x_{2} = 1\\ x_{3} = - \sqrt{3} \\ x_{4} = \sqrt{3} [/tex]

б)

[tex]2(x + 3) {}^{2} - 5(x + 3) + 2 = 0 \\ x + 3 = a \\ 2 {a}^{2} - 5a + 2 = 0 \\ D = ( - 5) {}^{2} - 4 \times 2 \times 2 = 25 - 16 = 9 \\ a _{1}= \frac{5 - 3}{2 \times 2} = \frac{2}{4} = 0.5 \\ a _{2}= \frac{5 + 3}{2 \times 2} = \frac{8}{4} = 2 \\ \\ 1) \: a = 0.5 \\ x + 3 = 0.5 \\ x = 0.5 - 3 \\ x_{1} = - 2.5 \\ \\ 2) \: a = 2 \\ x + 3 = 2 \\ x = 2 -3\\ x_{2} = -1[/tex]