Question

помогите пж срочно алгебра 8 клас​

Answer 1

$\displaystyle 1)\frac{x^2+x}{x^2+1}=0, x\in R\\\\x^2+x=0\\x(x+1)=0\\x_1=0,x_2=0\\\\2)\frac{3x-5}{2}=-1\\\\3x-5=-2\\3x=-2+5\\3x=3\\x=1\\\\3)\frac{2x^2-1}{x}=x,x\neq 0\\\\2x^2-1=x^2\\2x^2-x^2=1\\x^2=1\\x=б1$