Odpowiedź:
a)
[tex]x-\frac{5}{x}=0\\\\D=\mathbb{R}\setminus\lbrace0\rbrace\\\\\frac{x^2}{x}-\frac{5}{x}=0\\\frac{x^2-5}{x}=0\\x^2-5=0\\x^2=5\\x=\sqrt{5} \in D \ \vee \ x=-\sqrt{5} \in D[/tex]
b)
[tex]2x-\frac{1}{x}=3\\\\D=\mathbb{R}\setminus\lbrace0\rbrace\\\\\frac{2x^2}{x}-\frac{1}{x}-\frac{3x}{x}=0\\\frac{2x^2-1-3x}{x}=0\\2x^2-3x-1=0\\\Delta=9+8=17\\\sqrt{\Delta}=\sqrt{17}\\x=\frac{3-\sqrt{17}}{4}\in D \ \vee \ x=\frac{3+\sqrt{17}}{4}\in D[/tex]
c)
[tex]\frac{1}{x^2}+\frac{2}{x}=4\\\\D=\mathbb{R}\setminus\lbrace0\rbrace\\\\\frac{1}{x^2}+\frac{2x}{x^2}-\frac{4x^2}{x^2}=0\\\frac{1+2x-4x^2}{x^2}=0\\-4x^2+2x+1=0\\4x^2-2x-1=0\\\Delta=4+16=20\\\sqrt{\Delta}=2\sqrt{5}\\x=\frac{2-2\sqrt{5}}{8}=\frac{1-\sqrt{5}}{4}\in D \ \vee \ x=\frac{2+2\sqrt{5}}{8}=\frac{1+\sqrt{5}}{4} \in D[/tex]
d)
[tex]\frac{15}{x^2}-\frac{2}{x}-1=0\\\\D=\mathbb{R}\setminus\lbrace0\rbrace\\\\\frac{15}{x^2}-\frac{2x}{x^2}-\frac{x^2}{x^2}=0\\\frac{15-2x-x^2}{x^2}=0\\-x^2-2x+15=0\\x^2+2x-15=0\\\Delta=4+60=64\\\sqrt{\Delta}=8\\x=\frac{-2-8}{2}=-5\in D \ \vee \ x=\frac{-2+8}{2}=3\in D[/tex]
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Odpowiedź:
a)
[tex]x-\frac{5}{x}=0\\\\D=\mathbb{R}\setminus\lbrace0\rbrace\\\\\frac{x^2}{x}-\frac{5}{x}=0\\\frac{x^2-5}{x}=0\\x^2-5=0\\x^2=5\\x=\sqrt{5} \in D \ \vee \ x=-\sqrt{5} \in D[/tex]
b)
[tex]2x-\frac{1}{x}=3\\\\D=\mathbb{R}\setminus\lbrace0\rbrace\\\\\frac{2x^2}{x}-\frac{1}{x}-\frac{3x}{x}=0\\\frac{2x^2-1-3x}{x}=0\\2x^2-3x-1=0\\\Delta=9+8=17\\\sqrt{\Delta}=\sqrt{17}\\x=\frac{3-\sqrt{17}}{4}\in D \ \vee \ x=\frac{3+\sqrt{17}}{4}\in D[/tex]
c)
[tex]\frac{1}{x^2}+\frac{2}{x}=4\\\\D=\mathbb{R}\setminus\lbrace0\rbrace\\\\\frac{1}{x^2}+\frac{2x}{x^2}-\frac{4x^2}{x^2}=0\\\frac{1+2x-4x^2}{x^2}=0\\-4x^2+2x+1=0\\4x^2-2x-1=0\\\Delta=4+16=20\\\sqrt{\Delta}=2\sqrt{5}\\x=\frac{2-2\sqrt{5}}{8}=\frac{1-\sqrt{5}}{4}\in D \ \vee \ x=\frac{2+2\sqrt{5}}{8}=\frac{1+\sqrt{5}}{4} \in D[/tex]
d)
[tex]\frac{15}{x^2}-\frac{2}{x}-1=0\\\\D=\mathbb{R}\setminus\lbrace0\rbrace\\\\\frac{15}{x^2}-\frac{2x}{x^2}-\frac{x^2}{x^2}=0\\\frac{15-2x-x^2}{x^2}=0\\-x^2-2x+15=0\\x^2+2x-15=0\\\Delta=4+60=64\\\sqrt{\Delta}=8\\x=\frac{-2-8}{2}=-5\in D \ \vee \ x=\frac{-2+8}{2}=3\in D[/tex]