Помогите решить неравенства и систему неравенств,заранее спасибо.c подробным...

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\log_{0.2}(2x-3)\\ \left\lbrace\begin{array}{l} \log_{0.2}x(x-2)>\log_{0.2}(2x-3)\\ x>2 \end{array}\right.\\ \left\lbrace\begin{array}{l} x^2-2x<2x-3\\ x>2 \end{array}\right.\\ \left\lbrace\begin{array}{l} x^2-4x+3<0\\ x>2 \end{array}\right.\\ \boxed{x\in(2,3)}" alt="\log_{0.2}(x-2)+\log_{0.2}x>\log_{0.2}(2x-3)\\ \left\lbrace\begin{array}{l} \log_{0.2}x(x-2)>\log_{0.2}(2x-3)\\ x>2 \end{array}\right.\\ \left\lbrace\begin{array}{l} x^2-2x<2x-3\\ x>2 \end{array}\right.\\ \left\lbrace\begin{array}{l} x^2-4x+3<0\\ x>2 \end{array}\right.\\ \boxed{x\in(2,3)}" align="absmiddle" class="latex-formula">

$\dfrac{\sqrt{36-x^2}\cdot\log_{0.5}x}{x-2}\leqslant0\\ \left\lbrace\begin{array}{l} \left[\begin{array}{l} x^2=36\\ \dfrac{x-1}{x-2}\geqslant0 \end{array}\right.\\ x\in(0,2)\cup(2,6]\\ \end{array}$
$\boxed{x\in(0,1]\cup(2,6]}$

$\left\lbrace\begin{array}{lc} 3\sqrt{x+y}-2\sqrt{x-y}=4&(*)\\ 2\sqrt{x+y}-\sqrt{x-y}=3&(**) \end{array}\right.\\ 2(**)-(*)=\sqrt{x+y}=2\\ \sqrt{x-y}=2\cdot2-3=1\\ \left\lbrace\begin{array}{l} x+y=4\\ x-y=1 \end{array}\right.\\ (x,y)=(5/2,3/2)$

0\\ y^2+2x=29 \end{array}\right.\\ \left\lbrace\begin{array}{l} x^2-x-2=0\\ y^2=29-2x\\ x+y+1>0 \end{array}\right.\\ \left\lbrace\begin{array}{l} \left[\begin{array}{l} x=2\\ x=-1 \end{array}\right.\\ y^2=29-2x\\ x+y+1>0 \end{array}\right. " alt="\left\lbrace\begin{array}{l} 2^{\log_2(x+y+1)}=x^2+y-1\\ \log_{\sqrt{29}}(y^2+2x)=2 \end{array}\right.\\ \left\lbrace\begin{array}{l} x+y+1=x^2+y-1\\ x+y+1>0\\ y^2+2x=29 \end{array}\right.\\ \left\lbrace\begin{array}{l} x^2-x-2=0\\ y^2=29-2x\\ x+y+1>0 \end{array}\right.\\ \left\lbrace\begin{array}{l} \left[\begin{array}{l} x=2\\ x=-1 \end{array}\right.\\ y^2=29-2x\\ x+y+1>0 \end{array}\right. " align="absmiddle" class="latex-formula">

0 \end{array}\right.\\ \boxed{(x,y)\in\lbrace(2,5),(-1,\sqrt{31})\rbrace}" alt="\left\lbrace\begin{array}{l} \left[\begin{array}{l} x=2, y^2=25\\ x=-1, y^2=31 \end{array}\right.\\ x+y+1>0 \end{array}\right.\\ \boxed{(x,y)\in\lbrace(2,5),(-1,\sqrt{31})\rbrace}" align="absmiddle" class="latex-formula">

$\log_x(x^2+3)=\log_x(4x)\\ \left\lbrace\begin{array}{l} x^2+3=4x\\ x\in(0,1)\cup(1,\infty) \end{array}\right.\\ \left\lbrace\begin{array}{l} x^2-4x+3=0\\ x\in(0,1)\cup(1,\infty) \end{array}\right.\\ \boxed{x=3}$

nelle987_zn (148k баллов) 17 Март, 18