Решите неравенство [tex](\frac{1}{4} )^{\frac{x+1}{x-2} } \\[/tex] > [tex]64^{\frac{x-1}{x+2} }[/tex]. Created by oksan4ik4. algebra-ru

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August 2022 1 5 Report

Решите неравенство [tex](\frac{1}{4} )^{\frac{x+1}{x-2} } \\[/tex] > [tex]64^{\frac{x-1}{x+2} }[/tex]

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$\displaystyle\bf\\\Big(\frac{1}{4} \Big)^{\frac{x+1}{x-2} } >64^{\frac{x-1}{x+2} } \\\\\\\Big(4^{-1} \Big)^{\frac{x+1}{x-2} } >\Big(4^{3} \Big)^{\frac{x-1}{x+2} } \\\\\\4^{-\frac{x+1}{x-2} }> 4^{\frac{3x-3}{x+2} } \\\\\\\frac{x+1}{2-x} >\frac{3x-3}{x+2} \\\\\\\frac{x+1}{2-x} -\frac{3x-3}{x+2} >0\\\\\\\frac{(x+1)(x+2)-(3x-3)(2-x)}{(2-x)(x+2)} >0\\\\\\\frac{x^{2} +2x+x+2-6x+3x^{2} +6-3x}{(2-x)(x+2)} >0$

$\displaystyle\bf\\\frac{4x^{2} -6x+8}{(2-x)(x+2)} >0\\\\\\\frac{2x^{2} -3x+4}{(2-x)(x+2)} >0\\\\\\2x^{2} -3x+4=0\\\\D=(-3)^{2} -4\cdot 2 \cdot 4=9-32=-23\\\\\\D0 \ \Rightarrow \ \ 2x^{2} -3x+4>0 -$ при любых значениях x

+ + + + + (-2) - - - - - (2) + + + + +

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Ответ : x ∈ (- ∞ ; - 2) ∪ (2 ; + ∞)

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