Решите, что сможите. Даю 40 баллов.

$1)\; \; 4x^2-12=0\\\\4x^2=12\\\\x^2=3\; \; \Rightarrow \quad x=\sqrt3\; \; ili\; \; x=-\sqrt3\\\\2) \left \{ {{3x\ \textgreater \ 12+11x} \atop {5x-1\ \textless \ 0}} \right. \; \left \{ {{-12\ \textgreater \ 8x} \atop {5x\ \textless \ 1}} \right. \; \left \{ {{8x\ \textless \ -12} \atop {x\ \textless \ \frac{1}{5}}} \right. \; \left \{ {{x\ \textless \ -\frac{12}{8}} \atop {x\ \textless \ 0,2}} \right. \; \left \{ {{x\ \textless \ -1,5} \atop {x\ \textless \ 0,2}} \right. \; \Rightarrow \; x\ \textless \ -1,5\\\\3)\quad 3x^2+2x-1=0\\\\D=4+12=16\; ,\; \; x_1=\frac{-2-4}{6}=-1\; ,\; x_2=\frac{-2+4}{6}=\frac{1}{3}$

$3x^2+2x-1=3(x+1)(x-\frac{1}{3})=$ $(x+1)(3x-1)$

$4)\quad \frac{1}{x} = \frac{1}{a} + \frac{1}{b} \\\\\frac{1}{b}=\frac{1}{x}-\frac{1}{a}\\\\\frac{1}{b}=\frac{a-x}{ax}\\\\b=\frac{ax}{a-x}\\\\5)\; \; (sinx-cosx)^2+2\cdot sinx\cdot cosx=\\\\=sin^2x-2sinx\cdot cosx+cos^2x+2sinx\cdot cosx=\\\\=sin^2x+cos^2x=1$