Question

решите пожалуйста до г) дам много баллов срочно ​

Answer 1

$1)\frac{x^{2} }{2-x} =\frac{3x}{2-x}\\\\\frac{x^{2}}{2-x}-\frac{3x}{2-x}=0\\\\\frac{x^{2}-3x}{2-x}=0\\\\2-x\neq 0\Rightarrow x\neq 2\\\\x^{2}-3x=0\\\\x(x-3)=0\\\\x_{1}=0\\\\x-3=0\\\\x_{2}=3\\\\Otvet:\boxed{0;3}$

$2)\frac{2x^{2}+3x }{3-x}=\frac{x-x^{2} }{x-3} \\\\\frac{2x^{2}+3x}{3-x}+\frac{x-x^{2}}{3-x}=0\\\\\frac{2x^{2}+3x+x-x^{2}}{3-x}=0\\\\\frac{x^{2}+4x }{3-x}=0\\\\3-x\neq0\Rightarrow x\neq3\\\\x^{2}+4x=0\\\\x(x+4)=0\\\\x_{1}=0\\\\x+4=0\\\\x_{2} =-4\\\\Otvet:\boxed{-4;0}$

$3)\frac{5x-7}{x-3}=\frac{4x-3}{x} \\\\\frac{5x-7}{x-3}-\frac{4x-3}{x} =0\\\\\frac{5x^{2}-7x-4x^{2} +3x+12x-9 }{x(x-3)} =0\\\\\frac{x^{2}+8x-9 }{x(x-3)}=0\\\\x\neq 0;x\neq 3\\\\x^{2}+8x-9=0\\\\D=8^{2}-4*(-9)=64+36=100=10^{2}\\\\x_{1}=\frac{-8-10}{2}=\frac{-18}{2}=-9\\\\x_{2}=\frac{-8+10}{2}=1\\\\Otvet:\boxed{-9;1}$