\(x\sqrt[3]{\left(y+z\right).1.1}\le\dfrac{x\left(y+z+1+1\right)}{3}=\dfrac{xy+xz+2x}{3}\)

\(A\le\dfrac{xy+xz+2x+yz+yx+2y+xz+zy+2z}{3}=\dfrac{2\left(xy+yz+xz\right)+2\left(x+y+z\right)}{3}=\dfrac{2}{3}\left(xy+yz+xz\right)+1\le\dfrac{2}{3}.\dfrac{\left(x+y+z\right)^2}{3}+1=\dfrac{2}{3}.\dfrac{\left(\dfrac{3}{2}\right)^2}{3}+1=\dfrac{3}{2}\)

\(\Rightarrow\)\(\)\(Max=\dfrac{3}{2}\Leftrightarrow x=y=z=\dfrac{1}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=m\\\left(x+y\right)^2-2xy=2m+1\end{matrix}\right.\)

\(đặt:x+y=S;xy=P\Rightarrow\left\{{}\begin{matrix}S=m\\m^2-2S=2m+1\Leftrightarrow P=\dfrac{m^2-2m-1}{2}\end{matrix}\right.\)

\(\Rightarrow S^2\ge4P\Leftrightarrow m^2\ge2\left(m^2-2m-1\right)\Leftrightarrow2-\sqrt{6}\le m\le2+\sqrt{6}\)

\(1\) \(cách\) \(làm\) \(khác\) \(B=\dfrac{1}{\dfrac{x}{\sqrt{yz}}+2}+\dfrac{1}{\dfrac{y}{\sqrt{xz}}+2}+\dfrac{1}{\dfrac{z}{\sqrt{xy}}+2}\)

\(đặt:\left(\dfrac{x}{\sqrt{yz}};\dfrac{y}{\sqrt{xz}};\dfrac{z}{\sqrt{xy}}\right)=\left(a;b;c\right)\Rightarrow abc=1\)

\(\Rightarrow B=\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\)

\(ta\) \(đi\) \(cm:B\le1\Leftrightarrow\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\le1\)

\(\Leftrightarrow\left(b+2\right)\left(c+2\right)+\left(a+2\right)\left(c+2\right)+\left(a+2\right)\left(b+2\right)\le\left(a+2\right)\left(b+2\right)\left(c+2\right)\)

\(\Leftrightarrow ab+bc+ca+abc-4\ge0\Leftrightarrow ab+bc+ca\ge3\)

điều này đúng theo cosi \(ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}=3\)

\(dấu"="\Leftrightarrow a=b=c=1\Leftrightarrow x=y=z\)

\(\left\{{}\begin{matrix}x^2y^2-xy-2=0\left(1\right)\\x^2+y^2=x^2y^2\left(2\right)\end{matrix}\right.\)\(đặt:xy=t\Rightarrow\left(1\right)\Leftrightarrow t^2-t-2=0\Leftrightarrow\left[{}\begin{matrix}t=2=xy\\t=-1=xy\end{matrix}\right.\)

\(với:xy=2\Rightarrow\left(2\right)\Leftrightarrow x^2+y^2=4\Leftrightarrow\left(x+y\right)^2-2xy=4\Leftrightarrow\left(x+y\right)^2=8\Leftrightarrow x+y=\pm2\sqrt{2}\)

\(\Rightarrow\left\{{}\begin{matrix}xy=2\\x+y=\pm2\sqrt{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=y=\sqrt{2}\\x=y=-\sqrt{2}\end{matrix}\right.\)

\(với:xy=-1\Rightarrow\left(2\right)\) \(vô\) \(nghiệm\) \(do:VT,VP>0\)

\(\Rightarrow\left(x;y\right)=\left\{\pm\sqrt{2}\right\}\)

\(\sqrt{ab\left(a+c\right)\left(b+c\right)}\le\dfrac{\left(a+b\right)}{2}.\dfrac{\left(a+c+b+c\right)}{2}=\dfrac{\left(a+b\right)\left(a+b+2c\right)}{4}=\dfrac{a^2+b^2+2ab+2ac+2bc}{4}\le\dfrac{a^2+b^2+a^2+b^2+\dfrac{4a^2+c^2}{2}+\dfrac{4b^2+c^2}{2}}{4}=\dfrac{8a^2+8b^2+2c^2}{8}=\dfrac{4a^2+4b^2+c^2}{4}\)\(\Rightarrow\dfrac{9}{\sqrt{ab\left(a+c\right)\left(b+c\right)}}-\dfrac{32}{\sqrt{4+4a^2+4b^2+c^2}}\ge\dfrac{36}{4a^2+4b^2+c^2}-\dfrac{32}{\sqrt{4+4a^2+4b^2+c^2}}\)

\(đặt:\sqrt{4+4a^2+4b^2+c^2}=t\)

\(\Rightarrow\dfrac{9}{\sqrt{ab\left(a+c\right)\left(b+c\right)}}-\dfrac{32}{\sqrt{4+4a^2+4b^2+c^2}}\ge\dfrac{36}{t^2-4}-\dfrac{32}{t}\)

\(ta\) \(đi\) \(cminh:\dfrac{36}{t^2-4}-\dfrac{32}{t}\ge-5\Leftrightarrow36t-32\left(t^2-4\right)+5\left(t^3-4t\right)\ge0\Leftrightarrow\left(t-4\right)^2\left(5t+8\right)\ge0\left(đúng\right)\)

\(\Rightarrowđpcm:\Rightarrow dấu"="\Leftrightarrow\left\{{}\begin{matrix}a=b=\dfrac{c}{2}\\t=4=\sqrt{4a^2+4b^2+c^2+4}\end{matrix}\right.\)

\(\Leftrightarrow a=b=1;c=2\)

\(x^2+y^2+x+y+\dfrac{1}{2}=0\Leftrightarrow x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+y^2+\dfrac{2.1}{2}y+\dfrac{1}{2}=0\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2=0\)

\(;\left(x+\dfrac{1}{2}\right)^2\ge0;\left(y+\dfrac{1}{2}\right)^2\ge0\Rightarrow dấu"="\Leftrightarrow x=y=-\dfrac{1}{2}\)

\(\left(đk:x;y;z\ge0\right)hpt\Leftrightarrow\left\{{}\begin{matrix}\left(x-\sqrt{y}\right)^2=x-y\left(1\right)\\\left(y-\sqrt{z}\right)^2=y-z\left(2\right)\\\left(z-\sqrt{x}\right)^2=z-x\left(3\right)\end{matrix}\right.\)

\(\left(1\right)+\left(2\right)+\left(3\right)\Rightarrow0=\left(x-\sqrt{y}\right)^2+\left(y-\sqrt{z}\right)^2+\left(z-\sqrt{x}\right)^2\)

\(dấu"="\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{y}\\y=\sqrt{x}\\z=\sqrt{x}\end{matrix}\right.\)\(\Rightarrow hpt\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y}-y=0\\\sqrt{z}-z=0\\\sqrt{x}-x=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=y=z=1\\x=y=z=0\end{matrix}\right.\)

\(A=\dfrac{x^2+2x+2}{x^2-x+3}\Leftrightarrow A\left(x^2-x+3\right)=x^2+2x+2\)

\(\Leftrightarrow Ax^2-Ax+3A=x^2+2x+2\Leftrightarrow\left(A-1\right)x^2-x\left(A+2\right)+3A-2=0\left(1\right)\)

\(với:A=1\Rightarrow x=\dfrac{1}{3}\)

\(với:A\ne1\Rightarrow\Delta\ge0\Leftrightarrow\left(A+2\right)^2-4\left(A-1\right)\left(3A-2\right)\ge0\Leftrightarrow\dfrac{2}{11}\le A\le2\)

\(\Rightarrow gtnn=\dfrac{2}{11};gtln=2\)

\(đặt:\sqrt{x^2+y}=a;\sqrt{x^2+3}=b\left(a\ge0;b>0\right)\)\(\left(x^2+y\ge0\right)\)

\(\left\{{}\begin{matrix}\left(\sqrt{x^2+y}+\sqrt{x^2+3}\right)x=y-3\left(1\right)\\\sqrt{x^2+y}+\sqrt{x}=x+3\left(2\right)\end{matrix}\right.\)

\(\Rightarrow a^2-b^2=x^2+y-x^2-3=y-3\)

\(\Rightarrow pt\left(1\right)\Leftrightarrow\left(a+b\right)x=a^2-b^2=\left(a+b\right)\left(a-b\right)\)

\(\Leftrightarrow x\left(a+b\right)-\left(a+b\right)\left(a-b\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(x-a+b\right)=0\Leftrightarrow\left[{}\begin{matrix}a=b\Rightarrow\sqrt{x^2+y}=-\sqrt{x^2+3}< 0\left(loại\right)\\x=a-b\left(3\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow\sqrt{x^2+y}-3+\sqrt{x}-x=0\)

\(\Leftrightarrow a+\sqrt{a-b}=a-b+3\Leftrightarrow\sqrt{a-b}=3-b\Leftrightarrow\left\{{}\begin{matrix}b\le3\\a-b=\left(3-b\right)^2\end{matrix}\right.\)

\(\Rightarrow a-b=\left(3-b\right)^2\Rightarrow x=\left(3-b^2\right)\Rightarrow\sqrt{x^2+3}=b\Leftrightarrow\sqrt{\left(3-b\right)^4+3}=b\Leftrightarrow b=2\left(thỏa\right)\Rightarrow x=1\Rightarrow y=8\left(thỏa\right)\)

\(\)