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Question 1

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Question 2

$\frac{x^3-x^2-x+1}{x+1} + \frac{4}{x^2-2x+1} \leq 5\\\\ \frac{x^2(x-1)-(x-1)}{x+1} + \frac{4}{(x-1)^2} \leq 5\\\\ \frac{(x-1)(x^2-1)}{x+1} +\frac{4}{(x-1)^2} \leq 5\\\\ \frac{(x-1)(x-1)(x+1)}{x+1} + \frac{4}{(x-1)^2} -5 \leq 0\\\\(x-1)^2+\frac{4}{(x-1)^2}-5 \leq 0\\\\t=(x-1)^2\; ,\; \; \frac{t^2-5t+4}{t} \leq 0\; ,\; \; \frac{(t-1)(t-4)}{t} \leq 0\\\\Znaki:\; \; ---(0)+++(1)---(4)+++$

$t\in (-\infty ,0)\cup [\, 1,4\, ]\; \; \to$

$\left \{ {{(x-1)^2\ \textless \ 0\; net\; reshenij} \atop {1<=(x-1)^2<=4}} \right.$

$\left \{ {{(x-1)^2 \geq 1} \atop {(x-1)^2 \leq 4}} \right. \; \left \{ {{(x-1)^2-1 \geq 0} \atop {(x-1)^2-4 \leq 0}} \right. \; \left \{ {{(x-1-1)(x-1+1) \geq 0} \atop {(x-1-2)(x-1+2) \leq 0}} \right. \; \left \{ {{(x-2)x \geq 0} \atop {(x-3)(x+1) \leq 0}} \right. \\\\ \left \{ {{x \leq 0\; ili\; x \geq 2} \atop {-1 \leq x \leq 3}} \right. \; \; \Rightarrow \; \; x\in [-1;0\, ]\cup [\, 2;3\, ]\\\\ODZ:\; \; x\ne -1\; ;\; x\ne 1\\\\Otvet:\; \; (-1;0\, ]\cup [\, 2;3\, ]$

$2)\; \; \frac{\sqrt2cosx-1}{\sqrt{-5sinx}} =0\; \; \to \; \; \left \{ {{\sqrt2cosx-1=0} \atop {sinx\ \textless \ 0}} \right. \\\\ \left \{ {{cosx=\frac{1}{\sqrt2}=\frac{\sqrt2}{2}} \atop {-\pi +2\pi n\ \textless \ x\ \textless \ 2\pi n,\; n\in Z}} \right. \; \; \left \{ {{x=\pm \frac{\pi}{4}+2\pi k,\; k\in Z} \atop {-\pi +2\pi n\ \textless \ x\ \textless \ 2\pi n,\; n\in Z}} \right. \\\\x=-\frac{\pi}{4}+2\pi n,\; n\in Z\\\\x\in [\, \frac{3\pi }{2};3\pi \, ]\; \; \to \; \; \; x=\frac{7\pi}{4}$