[tex]Zad. 1\\\\a) \\\frac{(x-1)(x+3)}{x^2-1}=0\\x^2-1 \neq 0\\x^2 \neq 1\\x \neq 1 \text{ v } x \neq -1\\D\in R / \{-1, 1\}\\\\(x-1)(x+3)=0\\x-1=0\\x=1 \notin D\\\\x+3=0\\x=-3\\\\Odp. x=-3[/tex]
[tex]b) \\\\\frac{7x-11}{x+2}=2\\x+2 \neq 0\\x \neq -2\\D \in R / \{-2\}\\\\\frac{7x-11}{x+2}=2 /*x+2\\7x-11=2(x+2)\\7x-11=2x+4 \\7x-2x=4+11\\5x=15 /:5\\x=3\\\\Odp. x=3[/tex]
[tex]c) \\\\\frac{4}{3x+4}=5+x\\\\3x+4 \neq 0\\3x \neq -4 /:3\\x \neq -\frac43\\\\D\in R / \{-\frac43\}\\\\\frac{4}{3x+4}=5+x /*3x+4\\4=(5+x)(3x+4)\\4=15x+20+3x^2+4x\\0=3x^2+19x+16\\\Delta=(19)^2-4*3*16\\\Delta=361-192=169\\\sqrt{\Delta}=13\\x_1=\frac{-19-13}{6}=\frac{-32}6=-\frac{16}3\\x_2=\frac{-19+13}6=\frac{-6}6=-1[/tex]
[tex]d) \\\\\frac{3x+5}{x+3}=\frac{3}{2x-1}-1\\x+3 \neq 0\\x \neq -3\\\\2x-1 \neq 0 \\2x \neq 1 /:2\\x \neq \frac12\\\\D\in R / \{-3, \frac12\}[/tex]
[tex]\frac{3x+5}{x+3}=\frac{3}{2x-1}-1\\\frac{3x+5}{x+3}-\frac{3}{2x-1}=-1\\\frac{(3x+5)(2x-1)}{(x+3)(2x-1)}-\frac{3(x+3)}{(x+3)(2x-1)}=-1\\\frac{6x^2-3x+10x-5}{(x+3)(2x-1)}-\frac{3x+9}{(x+3)(2x-1)}=-1\\\frac{6x^2+7x-5-(3x+9)}{(x+3)(2x-1)}=-1 /*(x+3)(2x-1)\\6x^2+7x-5-3x-9=-(x+3)(2x-1)\\6x^2+4x-14=-(2x^2-x+6x-3)\\6x^2+4x-14=-2x^2-5x+3\\6x^2+2x^2+4x+5x-14-3=0\\8x^2+9x-17=0\\\Delta=9^2-4*8*(-17)=81+544=625\\[/tex]
[tex]\sqrt{\Delta}=25\\x_1=\frac{-9-25}{16}=\frac{-34}{16}=-\frac{17}8\\x_2=\frac{-9+25}{16}=\frac{16}{16}=1[/tex]
Zad. 2
a)
[tex]\frac{5}{x+2} \geq 3\\x+2 > 0\\x > -2\\\frac{5}{x+2} \geq 3 /*x+2\\5 \geq 3(x+2)\\5 \geq 3x+6\\5-6 \geq 3x\\3x \leq -1 /:3\\x \leq -\frac13\\x\in(-2; -\frac13 >[/tex]
b)
[tex]\frac{x^2+4x}{x^2-9} < 0\\(x+3)(x-3) < 0\\x+3 < 0\\x < -3\\x-3 < 0\\x < 3\\\\x^2+4x < 0\\\Delta=4^2=4\\\sqrt{\Delta}=4\\x_1=\frac{-4-4}2=\frac{-8}2=-4\\x_2=\frac{-4+4}2=0\\\\x\in < -4; -3) U < 0; 3)[/tex]
c)
[tex]\frac{5x}{x+3} < \frac{x+4}{2x-1}\\\frac{5x}{x+3}-\frac{x+4}{2x-1} < 0\\(x+3)(2x-1) < 0\\x+3 < 0\\x < -3\\2x-1 < 0\\2x < 1 /:2\\x < \frac12\\\\5x(2x-1)-(x+4)(x+3) < 0\\10x^2-5x-(x^2+3x+4x+12) < 0\\10x^2-5x-x^2-7x-12 < 0\\9x^2-12x-12 < 0\\\Delta=(-12)^2-4*9*(-12)=144+432=576\\\sqrt{\Delta}=24\\x_1=\frac{12-24}{18}=\frac{-12}{18}=-\frac{6}{9}=-\frac23\\x_2=\frac{12+24}{18}=2\\x\in(-3; -\frac23)U(\frac12; 2)[/tex]
