Question

помогите пожалуйста даю 35 баллов

Answer 1

$\displaystyle\bf\\1)\\\\\frac{3x+4}{x^{2} -16}=\frac{x^{2} }{x^{2} -16} \\\\\\\frac{3x+4}{x^{2} -16}-\frac{x^{2} }{x^{2} -16} =0\\\\\\\frac{3x+4-x^{2} }{x^{2} -16} =0\\\\\\\left \{ {{-x^{2} +3x+4=0} \atop {x^{2} -16\neq 0}} \right. \\\\\\\left \{ {{x^{2} -3x-4=0} \atop {x\neq -4 \ ; \ x\neq 4}} \right. \\\\\\x^{2} -3x-4=0\\\\Teorema \ Vieta:\\\\x_{1} =-1\\\\x_{2} =4-neyd\\\\Otvet: \ -1$

$\displaystyle\bf\\2)\\\\\frac{3}{x-5} +\frac{8}{x} =2\\\\\\\frac{3}{x-5} +\frac{8}{x} -2=0\\\\\\\frac{3\cdot x+8\cdot(x-5)-2\cdot x\cdot (x-5)}{x(x-5)} =0\\\\\\\frac{3x+8x-40-2x^{2} +10x}{x(x-5)} =0\\\\\\\frac{-2x^{2} +21x-40}{x(x-5)} =0\\\\\\\left \{ {{2x^{2} -21x+40=0} \atop {x\neq 0 \ ; \ x\neq 5}} \right. \\\\\\2x^{2} -21x+40=0\\\\D=(-21)^{2} -4\cdot 2\cdot 40=441-320=121=11^{2} \\\\\\x_{1} =\frac{21-11}{4} =2,5\\\\\\x_{2} =\frac{21+11}{4} =8\\\\\\Otvet: \ 2,5 \ ; \ 8$