\(f\left(x\right)=\left(m^2-1\right)x^2-8mx+9-m^2\ge0\)

\(TH1:m^2-1=0\Rightarrow m=\pm1\)

\(m=1\Rightarrow bpt\Leftrightarrow-8x+8\ge0\Leftrightarrow x\le1\left(loại\right)\)

\(m=-1\Rightarrow bpt\Leftrightarrow8x+8\ge0\Leftrightarrow x\ge-1\left(tm\right)\)

\(TH2:m^2-1>0\Leftrightarrow\left[{}\begin{matrix}m>1\\m< -1\end{matrix}\right.\)

\(\Delta=\left(8m\right)^2-4\left(9-m^2\right)\left(m^2-1\right)< 0\Leftrightarrow4x^4+24x^2+36=\left(2x^2\right)^2+2.2x^2.6+6^2=\left(2x^2+6\right)^2>0\left(\forall m\right)\)

\(\Rightarrow x1< x2\le0\Leftrightarrow\)\(\left\{{}\begin{matrix}x1x2\ge0\\x1+x2< 0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}9-m^2\ge0\\8m< 0\end{matrix}\right.\)

\(\Leftrightarrow-3\le m< 0\Rightarrow-3\le m< -1\)

\(TH3:m^2-1< 0\Leftrightarrow-1< m< 1\)

\(f\left(x\right)\ge0\Leftrightarrow\Delta\le0\Leftrightarrow\left(2x^2+6\right)^2\le0\left(vô-lí\right)\)

\(\Rightarrow-3\le m\le-1\)

\(đk:\left\{{}\begin{matrix}\Delta\ge0\\0< x1\le x2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5^2-4\left(-m^2+m+6\right)\ge0\\\left\{{}\begin{matrix}x1+x2>0\\x1x2>0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}4m^2-4m+1=\left(2m-1\right)^2\ge0\left(đúng\right)\\\left\{{}\begin{matrix}5>0đúng\\-m^2+m+6>0\Leftrightarrow-2< m< 3\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-2< m< 3\)

\(\Rightarrow\dfrac{1}{\sqrt{x1}}+\dfrac{1}{\sqrt{x2}}=\dfrac{3}{2}\Leftrightarrow\dfrac{\sqrt{x1}+\sqrt{x2}}{\sqrt{x1x2}}=\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{x1+x2+2\sqrt{x1x2}}{x1x2}=\dfrac{9}{4}\Leftrightarrow\dfrac{5+2\sqrt{-m^2+m+6}}{-m^2+m+6}=\dfrac{9}{4}\)

\(đặt::\sqrt{-m^2+m+6}=t\ge0\Rightarrow\dfrac{5+2t}{t^2}=\dfrac{9}{4}\)

\(\Rightarrow9t^2-8t-20=0\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-\dfrac{10}{9}\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{-m^2+m+6}=2\Leftrightarrow\left[{}\begin{matrix}m=2\left(tm\right)\\m=-1\left(tm\right)\end{matrix}\right.\)

\(B1:a;M=\dfrac{A}{B}=\dfrac{\dfrac{x+12}{\sqrt{x}-1}}{\dfrac{\sqrt{x}+2}{\sqrt{x}-1}}=\dfrac{x+12}{\sqrt{x}+2}=4+\dfrac{x-4\sqrt{x}+4}{\sqrt{x}+2}=4+\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}+2}\ge4\Rightarrow min=4\Leftrightarrow\sqrt{x}-2=0\Leftrightarrow x=4\left(tm\right)\)

\(b,\dfrac{A}{B}=\dfrac{\left(x+2\sqrt{x}\right)\sqrt{x}\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}+2\right)}=\dfrac{x\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\dfrac{x}{\sqrt{x}-1}>1\Leftrightarrow\dfrac{x}{\sqrt{x}-1}-1>0\Leftrightarrow\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\)

\(do:x-\sqrt{x}+1=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\Rightarrow\sqrt{x}-1>0\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)

\(3;M=\dfrac{B}{A}=\dfrac{x-1}{2\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)

\(\Rightarrow\dfrac{1}{m}-\dfrac{\sqrt{x}+1}{8}\ge1\Leftrightarrow\dfrac{2\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{8}-1\ge0\)

\(\Leftrightarrow\dfrac{16\sqrt{x}-\left(\sqrt{x}+1\right)^2-8\sqrt{x}-8}{8\sqrt{x}+8}\ge0\Rightarrow16\sqrt{x}-x-2\sqrt{x}-1-8\sqrt{x}-8\ge0\Leftrightarrow6\sqrt{x}-x-9\ge0\Leftrightarrow-\left(x-6\sqrt{x}+9\right)\ge0\Leftrightarrow\left(\sqrt{x}-3\right)^2\le0\Leftrightarrow x=9\left(tm\right)\)

\(\)\(d;đk:x\ge0;x\ne4\Rightarrow P=A.B=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)

\(\Rightarrow\left|\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\right|>\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)

\(TH1:\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\ge0\Leftrightarrow\sqrt{x}-1\ge0\Leftrightarrow x\ge1\left(x\ne4\right)\Rightarrow\)

\(P>P\left(vô-lý\right)\)

\(TH2:P< 0\Leftrightarrow\sqrt{x}-1< 0\Leftrightarrow0\le x< 1\)

\(\Rightarrow-P>P\Leftrightarrow-2P>0\Leftrightarrow P< 0\Leftrightarrow0\le x< 1\)

\(c17;f\left(x\right)=x^2-\left(m+2\right)x+4m+1\)

\(\Leftrightarrow\Delta>0\Leftrightarrow\left(m+2\right)^2-4\left(4m+1\right)=m^2-12m>0\Leftrightarrow\left[{}\begin{matrix}m< 0\\m>12\end{matrix}\right.\)

\(c18:\)\(x^2-2\left(m+1\right)x+m^2+3< 0\)

\(\Leftrightarrow\Delta'\le0\Leftrightarrow\left(m+1\right)^2-m^2-3\le0\Leftrightarrow m\le1\)

\(c19:\Leftrightarrow\Delta'\ge0\Leftrightarrow m^2-1\ge0\Leftrightarrow m^2\ge1\Leftrightarrow\left|m\right|\ge1\)

\(20.x^2+px+1=0\Leftrightarrow\left\{{}\begin{matrix}a+b=-p\\ab=1\end{matrix}\right.\)

\(x^2+qx+2=0\Leftrightarrow\left\{{}\begin{matrix}b+c=-q\\bc=2\end{matrix}\right.\)

\(\Rightarrow pq=\left(-p\right)\left(-q\right)=\left(a+b\right)\left(b+c\right)\)

\(\Rightarrow A=\left(a+b\right)\left(b+c\right)-\left(b-a\right)\left(b-c\right)=ab+ac+b^2+bc-b^2+bc+ab-ac=2ab+2bc=2+2.2=6\)

\(c19:\Rightarrow ADHE\) \(là\) \(hình\) \(chữ\) \(nhật\)(mình hướng dẫn thôi bn tự làm)

\(\Rightarrow\left\{{}\begin{matrix}AB^2+AC^2=100\\\dfrac{1}{4^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\end{matrix}\right.\)\(giải\) \(hệ\Rightarrow AB,AC\)

\(dùng:AB^2=BH.BC\Rightarrow BH=\dfrac{AB^2}{BC}=......\)

\(\Rightarrow CH=BC-BH=.....\)

\(\dfrac{1}{DH^2}=\dfrac{1}{BH^2}+\dfrac{1}{AH^2}\Rightarrow DH=....\)

\(\dfrac{1}{HE^2}=\dfrac{1}{AH^2}+\dfrac{1}{HC^2}\Rightarrow HE=,,,,,\Rightarrow S\left(ADHE\right)=DH.HE=....\)

\(a;gọi\) \(x_o\) \(là\) \(ngo\) \(chung\Rightarrow\left\{{}\begin{matrix}x_o^2+x_o+a=0\\x_o^2+a.x_o+1=0\end{matrix}\right.\)

\(đk:\left\{{}\begin{matrix}\Delta1\ge0\\\Delta2\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}1-4a\ge0\\a^2-4\ge0\end{matrix}\right.\)\(\Leftrightarrow a\le-2\)

\(\Leftrightarrow x_o+a-ax_o-1=0\Leftrightarrow x_o=1\Rightarrow a=-2\left(tm\right)\)

\(b,TH1:vô\) \(nghiệm\Leftrightarrow\left\{{}\begin{matrix}\Delta1< 0\\\Delta2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1-4a< 0\\a^2-4< 0\end{matrix}\right.\)\(\Leftrightarrow\dfrac{1}{4}< a< 2\)

\(ycđb\Leftrightarrow\left\{{}\begin{matrix}x1+x2=x1'+x2'\\x1.x2=x1'.x2'\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-1=-a\\a=1\end{matrix}\right.\)\(\Leftrightarrow a=1\left(tm\right)\)

\(x\ne0:đặt:x+\dfrac{1}{x}=t\)

\(pt\Leftrightarrow2t^2+4\left(t^2-2\right)^2-4\left(t^2-2\right)t^2=\left(x+4\right)^2\)

\(\Leftrightarrow2t^2+4\left(t^4-4t^2+4\right)-4\left(t^4-2t^2\right)=\left(x+4\right)^2\)

\(\Leftrightarrow2t^2+4t^4-16t^2+16-4t^4+8t^2=\left(x+4\right)^2\)

\(\Leftrightarrow-6t^2+16=\left(x+4\right)^2\)

\(\Leftrightarrow-6\left(x^2+2+\dfrac{1}{x^2}\right)+16=x^2+8x+16\)

\(\Leftrightarrow-6x^2-\dfrac{6}{x^2}-x^2-8x-12=0\Leftrightarrow-6x^4-x^4-8x^3-12x^2-6=0\Leftrightarrow-7x^4-8x^3-12x^2-6=0\left(vô-nghiệm\right)\)

(bn xem lại đề)

\(2.đk:x^2;y^2\ge2018\Leftrightarrow\left[{}\begin{matrix}x;y\le-\sqrt{2018}\\x;y\ge\sqrt{2018}\end{matrix}\right.\)

\(pt\Leftrightarrow\sqrt{x^2+11}-\sqrt{y^2+11}+\sqrt{x^2-2018}-\sqrt{y^2-2018}+x^2-y^2=0\)

\(\Leftrightarrow\left(x+y\right)\left(x-y\right)+\dfrac{x^2+11-y^2-11}{\sqrt{x^2+11}+\sqrt{y^2+11}}+\dfrac{x^2-2018-y^2+2018}{\sqrt{x^2-2018}+\sqrt{y^2-2018}}=0\)

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)\left[1+\dfrac{1}{\sqrt{x^2+11}+\sqrt{y^2+11}}+\dfrac{1}{\sqrt{x^2-2018}+\sqrt{y^2+2018}}>0\right]=0\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)

\(x=y\Rightarrow M=x^{11}-x^{2018}\)

\(x=-y\Rightarrow M=-y^{11}-y^{2018}=:vvv\) (đến đây chịu)

\(1.đk:\left(x+2\right)^3\ge0\Leftrightarrow x\ge-2\)

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

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

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

\(\Leftrightarrow x\left[\left(x-\sqrt{x+2}\right)\left(x+\sqrt{x+2}\right)\right]+2\left(x+2\right)\left(\sqrt{x+2}-x\right)=0\)

\(\Leftrightarrow\left(\sqrt{x+2}-x\right)\left[-x\left(\sqrt{x+2}+x\right)+2\left(x+2\right)\right]=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+2}=x\left(2\right)\\2\sqrt{x+2}=-x\left(3\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2=x+2\end{matrix}\right.\)\(\Leftrightarrow x=2\left(tm\right)\)

\(\left(3\right)\Leftrightarrow\left\{{}\begin{matrix}-x\ge0\Leftrightarrow x\le0\\x^2=4\left(x+2\right)\end{matrix}\right.\)\(\Leftrightarrow x=2-2\sqrt{3}\left(tm\right)\)

\(\left\{{}\begin{matrix}x=1-2t\\y=3+t\end{matrix}\right.\)\(\Rightarrow\overrightarrow{u}=\left(-2;1\right)\Rightarrow\overrightarrow{n}=\left(-1;-2\right)\)

\(\Rightarrow pt\) \(\Delta:-1\left(x-1\right)-2\left(y-3\right)=0\)

\(\Leftrightarrow-x+1-2y+6=0\Leftrightarrow-x-2y+7=0\Leftrightarrow y=\dfrac{-x}{2}+\dfrac{7}{2}\)

\(M\in\Delta\Rightarrow M=\left(x;-\dfrac{x}{2}+\dfrac{7}{2}\right)\)

\(\Rightarrow\overrightarrow{MA}=\left(1-x;\dfrac{x-3}{2}\right)\)

\(\Rightarrow\overrightarrow{MB}=\left(3-x;\dfrac{x-7}{2}\right);\Rightarrow\overrightarrow{MC}=\left(1-x;\dfrac{x-9}{2}\right)\)

\(\Rightarrow\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}=\left(5-3x;\dfrac{3x-19}{2}\right)\)

\(\Rightarrow\left|\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right|=\sqrt{\left(5-3x\right)^2+\left(\dfrac{3x-19}{2}\right)^2}=\sqrt{9x^2-30x+25+\dfrac{9x^2-114x+361}{4}}=\dfrac{\sqrt{36x^2-120x+100+9x^2-114x+361}}{2}=\dfrac{\sqrt{45x^2-264x+461}}{2}=\dfrac{\sqrt{45\left(x-\dfrac{44}{15}\right)^2+\dfrac{369}{5}}}{2}\ge\dfrac{\sqrt{\dfrac{365}{5}}}{2}=\dfrac{\sqrt{73}}{2}\Rightarrow min\Leftrightarrow x=\dfrac{44}{15}\Rightarrow M=\left(\dfrac{44}{15};\dfrac{61}{30}\right)\)

\(ktra\) \(tính\) \(toán\) \(lại\) \(giùm\)