[tex]\displaystyle\bf\\1)\\\\\frac{4x-10}{x-1} +\frac{x+6}{x+1} =4\\\\\\\frac{(4x-10)\cdot(x+1)+(x+6)\cdot(x-1)}{(x+1)(x-1)} -4=0\\\\\\\frac{4x^{2} +4x-10x-10+x^{2} -x+6x-6-4x^{2} +4}{(x+1)(x-1)} =0\\\\\\\frac{x^{2} -x-12}{(x+1)(x-1)} =0\\\\\\\left \{ {{x^{2} -x-12=0} \atop {x\neq -1 \ ; \ x\neq 1}} \right. \\\\\\x^{2} -x-12=0\\\\Teorema \ Vieta:\\\\x_{1} +x_{2} =1\\\\x_{1} \cdot x_{2} =-12\\\\x_{1} =4 \ \ ; \ \ x_{2} =-3\\\\Otvet:4 \ ; \ -3[/tex]
[tex]\displaystyle\bf\\2)\\\\\frac{1}{x} -\frac{10}{x^{2} -5x} =\frac{3-x}{x-5} \\\\\\\frac{1}{x} -\frac{10}{x(x-5)} -\frac{3-x}{x-5} =0\\\\\frac{x-5-10-(3-x)\cdot x}{x(x-5)} =0\\\\\\\frac{x-5-10-3x+x^{2} }{x(x-5)}=0\\\\\\\frac{x^{2} -2x-15}{x(x-5)} =0\\\\\\\left \{ {{x^{2} -2x-15=0} \atop {x\neq 0 \ \ ; \ \ x\neq 5}} \right. \\\\\\Teorema \ Vieta:\\\\x_{1} +x_{2} =2\\\\x_{1} \cdot x_{2}=-15\\\\x_{1} =-3 \ \ , \ \ x_{2} =5-neyd\\\\Otvet:-3 }[/tex]
Copyright © 2024 SCHOLAR.TIPS - All rights reserved.
Answers & Comments
[tex]\displaystyle\bf\\1)\\\\\frac{4x-10}{x-1} +\frac{x+6}{x+1} =4\\\\\\\frac{(4x-10)\cdot(x+1)+(x+6)\cdot(x-1)}{(x+1)(x-1)} -4=0\\\\\\\frac{4x^{2} +4x-10x-10+x^{2} -x+6x-6-4x^{2} +4}{(x+1)(x-1)} =0\\\\\\\frac{x^{2} -x-12}{(x+1)(x-1)} =0\\\\\\\left \{ {{x^{2} -x-12=0} \atop {x\neq -1 \ ; \ x\neq 1}} \right. \\\\\\x^{2} -x-12=0\\\\Teorema \ Vieta:\\\\x_{1} +x_{2} =1\\\\x_{1} \cdot x_{2} =-12\\\\x_{1} =4 \ \ ; \ \ x_{2} =-3\\\\Otvet:4 \ ; \ -3[/tex]
[tex]\displaystyle\bf\\2)\\\\\frac{1}{x} -\frac{10}{x^{2} -5x} =\frac{3-x}{x-5} \\\\\\\frac{1}{x} -\frac{10}{x(x-5)} -\frac{3-x}{x-5} =0\\\\\frac{x-5-10-(3-x)\cdot x}{x(x-5)} =0\\\\\\\frac{x-5-10-3x+x^{2} }{x(x-5)}=0\\\\\\\frac{x^{2} -2x-15}{x(x-5)} =0\\\\\\\left \{ {{x^{2} -2x-15=0} \atop {x\neq 0 \ \ ; \ \ x\neq 5}} \right. \\\\\\Teorema \ Vieta:\\\\x_{1} +x_{2} =2\\\\x_{1} \cdot x_{2}=-15\\\\x_{1} =-3 \ \ , \ \ x_{2} =5-neyd\\\\Otvet:-3 }[/tex]