Question

пожалуйста!!!!!!;?????!!!​

Answer 1

$1)\frac{3x}{x^{2}-2x+1 }-\frac{x+1}{x^{2}-x }=\frac{1}{x}\\\\\frac{3x}{(x-1)^{2}}-\frac{x+1}{x(x-1) }-\frac{1}{x}=0\\\\\frac{3x^{2}-x^{2}+1-x^{2}+2x-1}{x(x-1)^{2} }=0\\\\\frac{x^{2}+2x }{x(x-1)^{2}}=0\\\\\left \{ {{x^{2}+2x=0 } \atop {x\neq0 \ ; \ x-1\neq0}} \right. \\\\\left \{ {{x(x+2)=0} \atop {x\neq0 \ ; \ x\neq1}} \right.\\\\x_{1}=0-neyd\\\\x+2=0 \ \Rightarrow x_{2} =-2\\\\Otvet:\boxed{-2}$

$2)3a^{2}-5a-2=0\\\\D=(-5)^{2}-4*3*(-2)=25+24=49=7^{2}\\\\a_{1}=\frac{5-7}{6}=-\frac{1}{3}\\\\a_{2}=\frac{5+7}{6}=2\\\\3a^{2}-5a-2=3(a+\frac{1}{3})(a-2)=(3a+1)(a-2)\\\\\\\frac{3a^{2}-5a-2 }{a^{2}-4 }=\frac{(3a+1)(a-2)}{(a-2)(a+2)}=\boxed{\frac{3a+1}{a+2}} }$