Question

Решите пожалуйста дам 70 баллов

Answer 1

$1)\frac{x^{2}+3 }{x^{2}+1 }=2\\\\\frac{x^{2}+3 }{x^{2}+1 }-2=0\\\\\frac{x^{2}+3-2x^{2}-2}{x^{2}+1 }=0\\\\\frac{-x^{2}+1 }{x^{2} +1}=0\\\\\frac{x^{2}-1 }{x^{2}+1 }=0\\\\\left \{ {{x^{2}-1=0 } \atop {x^{2}+1\neq0}} \right. \\\\x^{2}=1\\\\\boxed{x_{1} =-1 \ ; \ x_{2} =1}\\\\\\2)\frac{3x^{2}-x }{1-x} =\frac{2}{1-x}\\\\\frac{3x^{2}-x }{1-x} -\frac{2}{1-x}=0\\\\\frac{3x^{2}-x-2 }{1-x}=0\\\\\left \{ {{3x^{2}-x-2=0 } \atop {1-x\neq 0}} \right.$

$\left \{ {{3x^{2}-x-2=0 } \atop {x\neq 1}} \right. \\\\3x^{2}-x-2=0\\\\D=(-1)^{2}-4*3*(-2)=1+24=25=5^{2}\\\\x_{1} =\frac{1-5}{6}=-\frac{2}{3}\\\\x_{2}=\frac{1+5}{6}=1-neyd\\\\Otvet:\boxed{-\frac{2}{3}}$

$3)\frac{2x+3}{x+2} =\frac{3x+2}{x}\\\\\frac{2x+3}{x+2} -\frac{3x+2}{x}=0\\\\\frac{2x^{2}+3x-3x^{2}-2x-6x-4}{x(x+2)}=0\\\\\frac{-x^{2}-5x-4 }{x(x+2)}=0\\\\\frac{x^{2}+5x+4 }{x(x+2)}=0\\\\\left \{ {{x^{2}+5x+4=0 } \atop {x\neq0 \ ; \ x+2\neq0}} \right.\\\\\left \{ {{x^{2}+5x+4=0 } \atop {x\neq0 \ ; \ x\neq-2}} \right.\\\\x^{2}+5x+4=0\\\\D=5^{2}-4*4=25-16=9=3^{2}\\\\x_{1} =\frac{-5-3}{2}=-4\\\\x_{2}=\frac{-5+3}{2}=-1\\\\Otvet:\boxed{-4 \ ; \ -1}$