Решите неравенство [tex] \frac{x}{x+3} - \frac{3}{x-1} + \frac{13}{ x^{2} +2x-3} \leq 0[/tex]. Created by Lenalook14. algebra-ru

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August 2022 2 6 Report

Решите неравенство
[tex] \frac{x}{x+3} - \frac{3}{x-1} + \frac{13}{ x^{2} +2x-3} \leq 0[/tex]

Verified answer

$\frac{x}{x+3} - \frac{3}{x-1} + \frac{13}{ x^{2}+2x-3 } \leq 0
\\\frac{x}{x+3} - \frac{3}{x-1} + \frac{13}{(x+3)(x-1)} \leq 0$
$\frac{ x^{2}-x-3x-9+13 }{(x+3)(x-1)} \leq 0$
$\frac{ x^{2}-4x+4 }{(x+3)(x-1)} \leq 0$
$\frac{(x-2) ^{2} }{(x+3)(x-1)} \leq 0$
______________________________
+ -3 - 1 + 2 +
x ∈ (-3 ;1) , 2

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ВикаВикторовна

Verified answer

ОДЗ: (х+3)(х-1)≠0; х≠ -3; 1
х(х-1)-3(х+3)+13 = 0
х²-х-3х-9+13 = 0
х²-4х+4 = 0
(х-2)² = 0
х-2 = 0
х = 2

+ - + +
------------₀-----------₀-------.----------->х
-3 1    2

х∈(-3; 1); {2}

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