Question

1. (1/x-1) + 2/x > (1/x+1)
2. (x+1)(x+2)(x+3)/(x-1)(x-2)(x-3)>1

Answer 1

$1)\; \; \frac{1}{x-1}+\frac{2}{x}>\frac{1}{x+1}\\\\\frac{x(x+1)+2(x-1)(x+1)-x(x-1)}{x(x-1)(x+1)}>0\\\\\frac{x^2+x+2x^2-2-x^2+x}{x(x-1)(x+1)}>0\\\\\frac{2(x^2+x-1)}{x(x-1)(x+1)} >0\\\\x^2+x-1=0\; ,\; \; D=1+4=5\; ,\; x_{1,2}=\frac{-1\pm \sqrt5}{2}\\\\x_1\approx -1,62\; ,\; \; x_2\approx 0,62\\\\---( -1,62)+++(-1)---(0)+++(0,62)---(1)+++\\\\x\in (\frac{-1-\sqrt5}{2},-1)\cup (0,\frac{-1+\sqrt5}{2})\cup (1,+\infty )$

$2)\; \; \frac{(x+1)(x+2)(x+3)}{(x-1)(x-2)(x-3)}>1\\\\\frac{(x+1)(x+2)(x+3)-(x-1)(x-2)(x-3)}{(x-1)(x-2)(x-3)}>0\\\\\frac{(x^2+3x+2)(x+3)-(x^2-3x+2)(x-3)}{(x-1)(x-2)(x-3)}>0\\\\\frac{x^3+3x^2+3x^2+9x+2x+6-(x^3-3x^2-3x^2+9x+2x-6)}{(x-1)(x-2)(x-3)}>0\\\\\frac{12x^2+12}{(x-1)(x-2)(x-3)}>0\; ,\; \; \frac{12(x^2+1)}{(x-1)(x-2)(x-3)}>0\\\\12(x^2+1)>0\; \; pri\; \; x\in (-\infty ,+\infty )\; \; \Rightarrow (x-1)(x-2)(x-3)>0\\\\---(1)+++(2)---(3)+++\\\\x\in (1,2)\cup (3,+\infty )$