\(\sqrt{x-2}+\sqrt{4-x}=m\le\sqrt{2\left(x-2+4-x\right)}=2\Rightarrow m\le2\)

\(m\ge\sqrt{x-2+4-x}=\sqrt{2}\Rightarrow m\ge\sqrt{2}\)

\(\Rightarrow m\in\left[\sqrt{2};2\right]\)

\(\left(đk:-1\le x\le3\right)\) 

\(pt\Leftrightarrow x\sqrt{x+1}+\sqrt{3-x}=2\sqrt{x^2+1}\)

\(VT\le\sqrt{\left(x^2+1\right)\left(x+1+3-x\right)}=2\sqrt{\left(x^2+1\right)}\left(bunhiacopxki\right)\)

\(\Rightarrow VT=VP\Leftrightarrow\dfrac{x}{\sqrt{x+1}}=\dfrac{1}{\sqrt{3-x}}\Leftrightarrow x\sqrt{3-x}=\sqrt{x+1}\Leftrightarrow x^2\left(3-x\right)=x+1\Leftrightarrow-\left(x-1\right)\left(x^2-2x-1\right)=0\Rightarrow x=....\)

\(1;\)

\(\dfrac{a^3}{b\left(c+2\right)}+\dfrac{b}{3}+\dfrac{c+2}{9}\ge3\sqrt[3]{\dfrac{a^3}{3.9}}=a\Rightarrow\dfrac{a^3}{b\left(c+2\right)}\ge a-\dfrac{b}{3}-\dfrac{c+2}{9}\)

\(tương\) \(tự\Rightarrow\dfrac{b^3}{c\left(a+2\right)}\ge b-\dfrac{c}{3}-\dfrac{a+2}{3};\dfrac{c^3}{a\left(b+2\right)}\ge c-\dfrac{a}{3}-\dfrac{b+2}{3}\)

\(\Rightarrow\Sigma\dfrac{a^3}{b\left(c+2\right)}\ge a+b+c-\left(\dfrac{2\left(a+b+c\right)}{3}\right)=\dfrac{a+b+c}{3}\ge\dfrac{3\sqrt[3]{abc}}{3}=1\)

\(dấu:="\Leftrightarrow a=b=c=1\)

\(d;pt\Leftrightarrow\left(\sqrt{x+5}-\sqrt{x+2}\right)\left(1+\sqrt{\left(x+2\right)\left(x+5\right)}\right)=3\)\(\left(1\right)\)\(\left(đkxđ:\left[{}\begin{matrix}x\ge-2\\x\le-5\end{matrix}\right.\right)\)

\(đặt:\)\(\sqrt{x+5}=a;\sqrt{x+2}=b\left(a;b\ge0\right)\Rightarrow a^2-b^2=3\)

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

\(\Leftrightarrow\left(a-b\right)\left(a-1\right)\left(b-1\right)=0\Leftrightarrow\left[{}\begin{matrix}a=b\\a=1\\b=1\end{matrix}\right.\)

\(với:a=b\Rightarrow\sqrt{x+5}=\sqrt{x+2}\Leftrightarrow x+5=x+2\Leftrightarrow3=0\left(loại\right)\) \(với:a=1\Rightarrow\sqrt{x+5}=1\Leftrightarrow x=-4\left(loại\right)\)

\(với:b=1\Rightarrow\sqrt{x+2}=1\Leftrightarrow x=-1\left(tm\right)\)

\(\left(x+y\right)xy=x^2+y^2-xy\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}=\dfrac{3}{4}\left(\dfrac{1}{x}-\dfrac{1}{y}\right)^2+\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\ge\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}\le4\)

\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\right)=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le4^2=16\)

\(đặt:\sqrt{x+1}+\sqrt{3-x}=t\left(2\le t\le2\sqrt[]{2}\right)\)

\(pt\Leftrightarrow\dfrac{2}{t}=1+\sqrt{\dfrac{t^2-4}{2}}\)

\(\Leftrightarrow\dfrac{2-t}{t}=\sqrt{\dfrac{t^2-4}{2}}\Rightarrow\left(2-t\right)^2=t^2\left(\dfrac{t^2-4}{2}\right)\)

\(\Rightarrow2\left(2-t\right)^2=t^4-4t^2\Leftrightarrow-\left(t-2\right)\left(t^3+2t^2-2t+4\right)=0\Leftrightarrow t=2\left(tm\right)\Rightarrow\sqrt{x+1}+\sqrt{3-x}=2\Rightarrow4+2\sqrt{\left(x+1\right)\left(3-x\right)}=4\Leftrightarrow\left(x+1\right)\left(3-x\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\) thử lại \(\Rightarrow\)\(x=3;x=-1\) \(là\) \(nghiệm\)

\(\Sigma\dfrac{x^2}{\sqrt{8x^2+3y^2+14xy}}=\Sigma\dfrac{x^2}{\sqrt{\left(4x+y\right)\left(2x+3y\right)}}\ge\dfrac{2\left(x+y+z\right)^2}{4x+y+2x+3y+4y+z+2y+3z+4z+x+2z+3x}=\dfrac{\left(x+y+z\right)^2}{5\left(x+y+z\right)}=\dfrac{x+y+z}{5}\left(đpcm\right)\)

\(\)\(M=4\left(a+b+c\right)^2+3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge4\left(a+b+c\right)^2+\dfrac{27}{a+b+c}=4\left(a+b+c\right)^2+\dfrac{13,5}{a+b+c}+\dfrac{13,5}{a+b+c}\ge3\sqrt[3]{4.13,5.13,5}=27\Rightarrow min=27\Leftrightarrow a=b=c=\dfrac{1}{2}\)

\(cminh:\)\(\dfrac{a}{a^3+a+1}\le\dfrac{4-a}{9}\left(1\right)\)

\(\left(1\right)\Leftrightarrow\)\(9a-\left(4-a\right)\left(a^3+a+1\right)\le0\Leftrightarrow\left(a-1\right)^2\left(a^2-2a-4\right)\le0\left(đúng\right)\Rightarrow\left(1\right)đúng\)

\(tương\) \(tự\Rightarrow\Sigma\dfrac{a}{a^3+a+1}\le\dfrac{4-a+4-b+4-c}{9}=\dfrac{-\left(a+b+c\right)+12}{9}\le\dfrac{-3\sqrt[3]{abc}+12}{9}=\dfrac{-3+12}{9}=1\left(đpcm\right)\)

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

\(\left(1\right)\Leftrightarrow\left(x-2\right)\left(x-y\right)=0\Leftrightarrow\left[{}\begin{matrix}x=y\Rightarrow\sqrt{2y}=y^2-2\left(3\right)\\x=2\Rightarrow\left(2\right)\Leftrightarrow\sqrt{2+y}=2y-2\left(4\right)\end{matrix}\right.\)

\(giải\left(3\right)và\left(4\right)\Rightarrow x;y\)