Question

Помогите кто-нибудь с 15, чего-то вообще не получается

Answer 1

$x\sqrt{x}+2\xqrt{x}+3 \leq \frac{6}{2-\sqrt{x}}\\\\x\sqrt{x}+2\sqrt{x}+3-\frac{6}{2-\sqrt{x}} \leq 0\\\\\frac{2x\sqrt{x}+4\sqrt{x}+6-x^2-2x-3\sqrt{x}-6}{2-\sqrt{x}} \leq 0\\\\\frac{2x\sqrt{x}-x^2-2x+\sqrt{x}}{2-\sqrt{x}} \leq 0$

Обозначим через  $t=\sqrt{x}>0$ .

$\frac{2t^3-t^4-2t^2+t}{2-t} \leq 0\; \; \; ,\; \; \; \frac{-t\cdot (t^3-2t^2+2t-1)}{-(t-2)} \leq 0\\\\\frac{t(t-1)(t^2-t+1)}{t-2} \leq 0\\\\t^2-t+1=0\; ,\; D=1-4=-3\ \textless \ 0\; \Rightarrow \; \; net\; kornej\; i\; \; \; t^2-t+1\ \textgreater \ 0\\\\---[0]+++[1]---(2)+++\\\\t\in (-\infty ,0\, ]\cup [\, 1,2)\\\\t\ \textgreater \ 0\; \; \Rightarrow \; \; \; t\in [\, 1,2)\; \; \Rightarrow \; \; 1 \leq t\ \textless \ 2\\\\1 \leq \sqrt{x}\ \textless \ 2\; \; \Rightarrow \; \; \left \{ {{\sqrt{x} \geq 1} \atop {\sqrt{x}\ \textless \ 2}} \right.$

$\left \{ {{x \geq 1} \atop {0<= x\ \textless \ 4}} \right. \; ;\; \;x\in [\, 1,4)$