Решите неравенство: [tex]x\sqrt{x^{2}-x-2}\geq0[/tex]. Created by Student59. algebra-ru

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Student59 @Student59

August 2022 1 29 Report

Решите неравенство:

[tex]x\sqrt{x^{2}-x-2}\geq0[/tex]

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Nik133

$x\sqrt{x^{2}-x-2}\geq0 \\ x\geq0 \\ x^2-x-2\geq0 \\ x^2-x-2=0 \\ D=1+2*4=9=3^2 \\ x_1=\frac{1+3}{2}=2 \\ \\ x_2=\frac{1-3}{2}=-1 \\ \\ x \in [2; +\infty)$

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