\(1.A=\sqrt{\sqrt{3}-\sqrt{1-\sqrt{12-2.\sqrt{12}.\sqrt{9}+9}}}=\sqrt{\sqrt{3}-\sqrt{1-\sqrt{\left(\sqrt{12}-\sqrt{9}\right)^2}}}=\sqrt{\sqrt{3}-\sqrt{1-\sqrt{12}+\sqrt{9}}}=\sqrt{\sqrt{3}-\sqrt{3-2\sqrt{3}+1}}=\sqrt{\sqrt{3}-\sqrt{\left(\sqrt{3}-1\right)^2}}=1\)
\(2.a;đk:\forall x;đặt:x^2=t\ge0\Rightarrow\dfrac{3}{t^2+t+1}+5=3t\left(t+1\right)\)
\(\Rightarrow\dfrac{3+5t^2+5t+5}{t^2+t+1}=3t\left(t+1\right)\Leftrightarrow5t^2+5t+8=3t\left(t+1\right)\left(t^2+t+1\right)\)
\(\Leftrightarrow\left(t-1\right)\left(t+2\right)\left(3t^2+3t+4>0\right)=0\Leftrightarrow\left[{}\begin{matrix}t=1\left(tm\right)\Rightarrow x=\pm1\\t=-2\left(loại\right)\end{matrix}\right.\)
\(b;\left\{{}\begin{matrix}2x^2-3xy+y^2=0\Leftrightarrow\left(x-y\right)\left(2x-y\right)=0\\3x^2+2xy-y^2-3x+y=2\left(3\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=y\left(1\right)\\y=2x\left(2\right)\end{matrix}\right.\\3x^2+2xy-y^2-3x+y=2\left(3\right)\end{matrix}\right.\)
\(thế\left(1\right)và\left(2\right)\) \(vào\left(3\right)\Rightarrow\left(x;y\right)\)
\(B2;a;mx+y=2\Leftrightarrow y=2-mx\)
\(\Rightarrow\dfrac{x^2}{2}=2-mx\Leftrightarrow x^2+2mx-4=0\Rightarrow\Delta'=m^2+4>0\left(\forall m\right)\Rightarrowđpcm\)
\(b;\Rightarrow\left\{{}\begin{matrix}x1+x2=-2m\\x1.x2=-4\end{matrix}\right.\) \(\Rightarrow A\left(x1;y1\right);B\left(x2;y2\right)\Leftrightarrow A\left(x1;2-mx1\right);B\left(x2;2-mx2\right)\)
\(\Rightarrow AB=\sqrt{\left(x1-x2\right)^2+\left(2-mx1-2+mx2\right)^2}=\sqrt{\left(x1-x2\right)^2+\left(mx1-mx2\right)^2}=\sqrt{\left(m^2+1\right)\left(x1-x2\right)^2}=\sqrt{\left(m^2+1\right)\left[\left(x1+x2\right)^2-4x1x2\right]}=\sqrt{\left(m^2+1\right)\left(4m^2+16\right)}\ge\sqrt{16}=4\Leftrightarrow m=0\)
\(b;;2\left(x-2\right)=\sqrt{2\left(x^2-x-1\right)}\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\4\left(x-2\right)^2=2\left(x^2-x-1\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\\left[{}\begin{matrix}x=\dfrac{1}{2}\left(7-\sqrt{13}\right)\\x=\dfrac{1}{2}\left(7+\sqrt{13}\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{1}{2}\left(7+\sqrt{13}\right)\Rightarrow T=5\)
\(B3..\)
\(\dfrac{4b^2}{a-1}+\dfrac{5b^2}{b-2}=4\left(a-1\right)+\dfrac{4}{a-1}+8+5\left(b-2\right)+\dfrac{20}{b-2}+20\ge2\sqrt{4^2}+8+2\sqrt{5.20}+20=56\left(đpcm\right)\)
\(a,2x^2-\left(m+2\right)x-7+m^2=0\)
\(yêu\) \(cầu\Leftrightarrow\left\{{}\begin{matrix}2\left(-7+m^2\right)< 0\left(1\right)\\\left|x1\right|=\dfrac{1}{x2}\left(2\right)\\m>0\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}m< -\sqrt{7}\\m>\sqrt{7}\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow-x1=\dfrac{1}{x2}\left(do:x1< 0\right)\Leftrightarrow x1.x2=-1\Leftrightarrow\dfrac{-7+m^2}{2}=-1\Leftrightarrow m^2=5\Leftrightarrow m=\pm\sqrt{5}\left(ktm\right)\Rightarrow m=\varnothing\)
\(c;x^2-xy+y^2=2\Leftrightarrow2x^2-2xy+2y^2-4=0\Leftrightarrow\left(x-y\right)^2+x^2+y^2-4=0\Leftrightarrow\left(x-y\right)^2=4-\left(x^2+y^2\right)\ge0\Rightarrow x^2+y^2\le4\)
\(\Rightarrow P\le4+xy=4+x^2+y^2-2=2+x^2+y^2\le2+4=6\Rightarrow maxP=6\)
\(dấu"="xayra\Leftrightarrow\left\{{}\begin{matrix}x=y\\x^2-xy+y^2=2\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)\)
\(..x^2-xy+y^2=2\Leftrightarrow\left(X+y\right)^2=2+3xy\ge0\Leftrightarrow xy\ge-\dfrac{2}{3}\Rightarrow P\ge x^2+y^2-\dfrac{2}{3}=2+xy-\dfrac{2}{3}\ge2-\dfrac{2}{3}-\dfrac{2}{3}=\dfrac{2}{3}\Rightarrow min=\dfrac{2}{3}\Leftrightarrow\left\{{}\begin{matrix}x=-y\\x^2-xy+y^2=2\end{matrix}\right.\)
