\(\Leftrightarrow9\le x^2\le34\Leftrightarrow-81\le x\le1156\)

\(=\dfrac{-x^2-2x+3+x^2+x}{\left(x-3\right)\left(x+3\right)}=\dfrac{-x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{-1}{x+3}\)

\(=x\left(1-4x^2\right)=x\left(1-2x\right)\left(1+2x\right)\)

\(a,=1-2x+x^2+2x-x^2=1\\ b,=\dfrac{\left(x+2\right)^2}{x+2}=x+2\)

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{a+b+c}=1\\ \Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Rightarrow a=b=c\)

\(2,\)

Ta có \(\left(y-2x\right)^2=\left(-2x+y\right)^2=\left[\dfrac{1}{3}\left(-6x\right)+\dfrac{1}{4}\left(4y\right)\right]^2\)

\(\Leftrightarrow\left(y-2x\right)^2\le\left[\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{4}\right)^2\right]\left[\left(-6x\right)^2+\left(4y\right)^2\right]=\dfrac{5^2}{3^2\cdot4^2}\left(36x^2+16y^2\right)=\dfrac{5^2}{4^2}\\ \Leftrightarrow\left|y-2x\right|\le\dfrac{5}{4}\\ \Leftrightarrow-\dfrac{5}{4}\le y-2x\le\dfrac{5}{4}\\ \Leftrightarrow\dfrac{15}{4}\le y-2x+5\le\dfrac{25}{4}\)

\(Max\Leftrightarrow\left\{{}\begin{matrix}-18x=16y\\y-2x=\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{5}\\y=\dfrac{9}{20}\end{matrix}\right.\\ Min\Leftrightarrow\left\{{}\begin{matrix}-18x=16y\\y-2x=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-\dfrac{9}{20}\end{matrix}\right.\)

làm đầy đủ vào chứ? min đâu

\(ĐK:-4\le x\le\dfrac{1}{2}\\ PT\Leftrightarrow\left(\sqrt{x+4}-2\right)-\left(\sqrt{1-x}-1\right)-\left(\sqrt{1-2x}-1\right)=0\\ \Leftrightarrow\dfrac{x}{\sqrt{x+4}+2}+\dfrac{x}{\sqrt{1-x}+1}+\dfrac{2x}{\sqrt{1-2x}+1}=0\\ \Leftrightarrow x\left(\dfrac{1}{\sqrt{x+4}+2}+\dfrac{1}{\sqrt{1-x}+1}+\dfrac{2}{\sqrt{1-2x}+1}\right)=0\)

Ta thấy ngoặc lớn >0 với mọi x

\(\Rightarrow x=0\left(tm\right)\)

\(1,\\ a,A\le\sqrt{\left(x-3+7-x\right)\left(1+1\right)}=\sqrt{8}=2\sqrt{2}\\ A^2=4+2\sqrt{\left(x-3\right)\left(7-x\right)}\ge4\Leftrightarrow A\ge2\\ \Leftrightarrow2\le A\le2\sqrt{2}\\ \left\{{}\begin{matrix}A_{min}\Leftrightarrow\left(x-3\right)\left(7-x\right)=0\Leftrightarrow...\\A_{max}\Leftrightarrow x-3=7-x\Leftrightarrow x=5\end{matrix}\right.\)

\(B=\dfrac{\dfrac{5}{2}\left(4x^4+4x^2+1\right)+2\left(x^4-x^2+\dfrac{1}{4}\right)}{\left(2x^2+1\right)^2}\\ B=\dfrac{\dfrac{5}{2}\left(2x^2+1\right)^2+2\left(x^2-\dfrac{1}{2}\right)^2}{\left(2x^2+1\right)^2}=\dfrac{5}{2}+\dfrac{2\left(x^2-\dfrac{1}{2}\right)^2}{\left(2x^2+1\right)^2}\ge\dfrac{5}{2}\)

\(B=\dfrac{3\left(4x^4+4x^2+1\right)-4x^2}{\left(1+2x^2\right)^2}=\dfrac{3\left(1+2x^2\right)^2-4x^2}{\left(1+2x^2\right)^2}=3-\dfrac{4x^2}{\left(1+2x^2\right)^2}\)

Vì \(-\dfrac{4x^2}{\left(1+2x^2\right)^2}\le0\Leftrightarrow B\le3\)

\(\Leftrightarrow\left\{{}\begin{matrix}B_{min}\Leftrightarrow x^2=\dfrac{1}{2}\Leftrightarrow x=\pm\dfrac{1}{\sqrt{2}}\\B_{max}\Leftrightarrow x=0\end{matrix}\right.\)

\(A=\dfrac{x+2}{x^2-x+3}\Leftrightarrow Ax^2-Ax+3A=x+2\\ \Leftrightarrow Ax^2-x\left(A+1\right)+3A-2=0\\ \Leftrightarrow\Delta=\left(A+1\right)^2-4A\left(3A-2\right)\ge0\\ \Leftrightarrow-11A+10A+1\ge0\\ \Leftrightarrow-\dfrac{1}{11}\le A\le1\)

Mà \(A\in Z\Leftrightarrow A\in\left\{0;1\right\}\)

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

Vậy \(x\in\left\{-2;1\right\}\Leftrightarrow A\in Z\)