Решите пожалуйста не получается у меня

Решите задачу:

$1)\; \; (x-2)^2\ \textless \ \sqrt3(x-2)\\\\(x-2)^2-\sqrt3(x-2)\ \textless \ 0\\\\(x-2)(x-2-\sqrt3)\ \textless \ 0\\\\x-2=0\; \; \to \; \; x=2\\\\x-2-\sqrt3=0\; \; \to \; \; \; x=2+\sqrt3\\\\+++(2)---(2+\sqrt3)+++\\\\x\in (2;\, 2+\sqrt3)\\\\\\2)\; \; x^2\cdot (-x^2-100) \leq 100\cdot (-x^2-100)\\\\x^2\cdot \underbrace {(-x^2-100)}-100\cdot \underbrace {(-x^2-100)} \leq 0\\\\\underbrace {(-x^2-100)}\cdot (x^2-100) \leq 0\\\\-(x^2+100)\cdot (x-10)(x+10) \leq 0\; |\cdot (-1)\\\\\underbrace {(x^2+100)}_{\ \textgreater \ 0}\cdot (x-10)(x+10) \geq 0$

$+++[-10\, ]---[\, 10\, ]+++\\\\x\in (-\infty ,-10\, ]\cup [\, 10,+\infty )\\\\\\3)\; \; \frac{1}{(x-3)^2} - \frac{3}{x-3} -4=0\; ,\; \; ODZ:\; \; x\ne 3\\\\ \frac{1-3(x-3)-4(x-3)^2}{(x-3)^2} =0\\\\ \frac{1-3x+9-4x^2+24x-36}{(x-3)^2} =0\\\\ \frac{-4x^2+21x-26}{(x-3)^2} =0\; \; \to \; \; 4x^2-21x+26=0\\\\D=25\; ,\; x_1=\frac{21-5}{8}=2\; ,\; \; x_2=\frac{21+5}{8}=\frac{13}{2}=6,5\\\\Otvet:\; \; x_1=2,\; \; x_2=6,5\; .$