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July 2022 2 35 Report

Решите неравенство (оно несложное, но я что-то туплю):
[tex]\sqrt{x+ 2} - \sqrt{3x - 1} \ \textgreater \ \sqrt{x - 1 }[/tex]

NeZeRAvix

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Решение в скане......

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NNNLLL54

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$\sqrt{x+2}-\sqrt{3x-1}>\sqrt{x-1}\\\\ODZ:\; \; \left \{ {{x\geq -2\; ,\; x\geq 1/3} \atop {x\geq 1}} \right. \; \; \to \; \; x\geq 1\\\\\underbrace {\sqrt{x+2}}_{\geq 0}>\underbrace {\sqrt{x-1}+\sqrt{3x-1}}_{\geq 0}\\\\x+2>(x-1)+2\cdot \sqrt{x-1}\cdot \sqrt{3x-1}+(3x-1)\\\\x+2>4x-2+2\cdot \sqrt{(x-1)(3x-1)}\\\\2\cdot \sqrt{(x-1)(3x-1)}$

$x_1=\frac{-4-2\sqrt{13}}{3}\approx -3,74\; \; ;\; \; x_2=\frac{-4+2\sqrt{13}}{3}\approx 1,07\\\\(x-\frac{-4-2\sqrt{13}}{3})(x-\frac{-4+2\sqrt{13}}{3})$

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