Question

Знайти корені рівняння

Answer 1

$\displaystyle\bf\\\frac{3x-5}{x^{2} -1} =\frac{3x+2}{x^{2} +x} -\frac{6x-5}{x^{2} -x} \\\\\\\frac{3x-5}{(x -1)(x+1)} -\frac{3x+2}{x\cdot(x+1)} +\frac{6x-5}{x\cdot(x-1)} =0\\\\\\\frac{(3x-5)\cdot x-(3x+2)\cdot(x-1)+(6x-5)\cdot(x+1)}{x(x-1)(x+1)} =0\\\\\\\frac{3x^{2} -5x-3x^{2} +3x-2x+2+6x^{2} +6x-5x-5}{x(x-1)(x+1)}=0\\\\\\\frac{6x^{2} -3x-3}{x(x-1)(x+1)} =0\\\\\\\frac{3\cdot(2x^{2} -x-1)}{x(x-1)(x+1)} =0\\\\\\\frac{2x^{2} -x-1}{x(x-1)(x+1)} =0$

$\displaystyle\bf\\\left \{ {{2x^{2} -x-1=0} \atop {x\neq 0 \ , \ x-1\neq 0 \ , \ x+1\neq 0}} \right. \\\\\\\left \{ {{2x^{2} -x-1=0} \atop {x\neq 0 \ , \ x\neq 1 \ , \ x\neq -1}} \right. \\\\\\2x^{2} -x-1=0\\\\D=(-1)^{2} -4\cdot 2\cdot (-1)=1+8=9=3^{2} \\\\\\x_{1} =\frac{1-3}{4} =\frac{-2}{4} =-0,5\\\\\\x_{2} =\frac{1+3}{4} =\frac{4}{4} =1- \ neyd\\\\\\Otvet \ : \ -0,5$