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June 2022 1 7 Report

Найдите произведение большего корня на количество корней уравнения (с решением!!!):
1). [tex] \sqrt[5]{ \frac{16x}{x-1} } + \sqrt[5]{ \frac{x-1}{16x} } = \frac{5}{2} [/tex]

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arsenlevadniy

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$\sqrt[5]{\frac{16x}{x-1}}+\sqrt[5]{\frac{x-1}{16x}}=\frac{5}{2}, \\ \left \{ {{x-1\neq0,} \atop {16x\neq0;}} \right. \left \{ {{x\neq1,} \atop {x\neq0;}} \right. \\ \sqrt[5]{\frac{16x}{x-1}}=t, \\ \sqrt[5]{\frac{x-1}{16x}}=(\sqrt[5]{\frac{16x}{x-1}})^{-1}=\frac{1}{t}, \\ t+\frac{1}{t}=\frac{5}{2}, \\ 2t^2+2=5t, \\ 2t^2-5t+2=0, \\ D=9, \\ t_1=\frac{1}{2}, t_2=2, \\$
$\sqrt[5]{\frac{16x}{x-1}}=\frac{1}{2}, \\ \frac{16x}{x-1}=\frac{1}{2^5}, \\ \frac{16x}{x-1}-\frac{1}{32}=0, \\ \frac{512x}{x-1}-1=0, \\ \frac{512x-x+1}{x-1}=0, \\ 511x+1=0, \\ x_1=-\frac{1}{511}; \\$
$\sqrt[5]{\frac{16x}{x-1}}=2, \\ \frac{16x}{x-1}=2^5, \\ \frac{16x}{x-1}=32, \\ \frac{16x}{x-1}-32=0, \\ \frac{16x-32x+32}{x-1}=0, \\ -16x+32=0, \\ x_2=2; \\ \\ 2\cdot2=4$

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