\(1,\Leftrightarrow x^2-2x-3x+6=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\\ 2,\Leftrightarrow x^2-3x-4x+12=0\\ \Leftrightarrow\left(x-3\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\\ 3,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+2\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\\ 4,\Leftrightarrow x^2+3x+4x+12=0\\ \Leftrightarrow\left(x+3\right)\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-3\end{matrix}\right.\\ 5,\Leftrightarrow x^2-4x+3x-12=0\\ \Leftrightarrow\left(x-4\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

\(6,\Leftrightarrow2x^2+x-4x-2=0\\ \Leftrightarrow\left(2x+1\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\\ 7,\Leftrightarrow3x^2+3x-2x-2=0\\ \Leftrightarrow\left(x+1\right)\left(3x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{2}{3}\end{matrix}\right.\\ 8,\Leftrightarrow4x^2-8x+x-2=0\\ \Leftrightarrow\left(x-2\right)\left(4x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{4}\end{matrix}\right.\\ 9,\Leftrightarrow4x^2-3x+8x-6=0\\ \Leftrightarrow\left(4x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-2\end{matrix}\right.\\ 10,\Leftrightarrow3x^2+9x+x+3=0\\ \Leftrightarrow\left(x+3\right)\left(3x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{3}\end{matrix}\right.\)

\(a,ĐK:x\ne0;x-1\ne0\Leftrightarrow x\ne0;x\ne1\\ b,ĐK:x^2-4=\left(x-2\right)\left(x+2\right)\ne0\Leftrightarrow x\ne2;x\ne-2\)

\(a,\Rightarrow8x^2+2x+28x+7=0\\ \Rightarrow2x\left(4x+1\right)+7\left(4x+1\right)=0\\ \Rightarrow\left(2x+7\right)\left(4x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\\ b,Sửa:x^3-7x-6=0\\ \Rightarrow x^3-x-6x-6=0\\ \Rightarrow x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x^2-x-6\right)=0\\ \Rightarrow\left(x+1\right)\left(x^2-3x+2x-6\right)=0\\ \Rightarrow\left(x+1\right)\left(x-3\right)\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=3\\x=-2\end{matrix}\right.\)

Chu vi là \(4,2\cdot3,14=13,188\left(dm\right)\)

Bán kính là \(4,2:2=2,1\left(dm\right)\)

Diện tích là \(2,1\cdot2,1\cdot3,14=13,8474\left(dm^2\right)\)

Vì AM=MB và OA=OB nên OM là trung trực AB

Do đó OM\(\bot\)AB tại H

Áp dụng HTL: \(OA^2=OH\cdot OM\Rightarrow OH=\dfrac{OA^2}{OM}=\dfrac{3^2}{5}=1,8\left(cm\right)\)

\(\Rightarrow HM=OM-OH=5-1,8=3,2\left(cm\right)\)

Kẻ \(MI\text{//}AC;DH\bot MN\left(H\in MN\right);IK\bot MN\left(K\in MN\right)\)

\(DHKI\) là hcn \(\Rightarrow DH=IK\Rightarrow S_{DMN}=S_{IMN}\)

Ta có \(\left\{{}\begin{matrix}\Delta AMN\sim\Delta ABC\\\Delta BMI\sim\Delta ABC\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{S_{AMN}}{S_{ABC}}=\left(\dfrac{AM}{AB}\right)^2\\\dfrac{S_{BMI}}{S_{ABC}}=\left(\dfrac{BM}{AB}\right)^2\end{matrix}\right.\)

\(\Rightarrow\dfrac{S_{AMN}+S_{BMI}}{S_{AB}}=\dfrac{AM^2+BM^2}{AB^2}\ge\dfrac{\dfrac{1}{2}\left(AM+MB\right)^2}{AB^2}\)

\(\Rightarrow\dfrac{S_{ABC}-S_{MNCI}}{S_{ABC}}\ge\dfrac{1}{2}\\ \Rightarrow1-\dfrac{S_{MNCI}}{S_{ABC}}\ge\dfrac{1}{2}\Rightarrow\dfrac{S_{MNCI}}{S_{ABC}}\le\dfrac{1}{2}\\ \Rightarrow S_{MNCI}\le\dfrac{1}{2}S_{ABC}\\ \Rightarrow2\cdot S_{DMN}\le\dfrac{1}{2}S_{ABC}\\ \Rightarrow S_{DMN}\le\dfrac{1}{4}S_{ABC}\)

Dấu \("="\Leftrightarrow AM=MB\Leftrightarrow M\) là trung điểm \(AB\Leftrightarrow N\) là trung điểm AC

Khi đó d đi qua trung điểm AB và AC

\(\Delta ADB\text{ cân tại A}\Rightarrow\widehat{ADB}=\dfrac{180^0-\widehat{BAD}}{2}=65^0\\ \text{Ta có }\widehat{MBH}=\widehat{BCD}=\widehat{ADN}=\widehat{BAD}=50^0\\ \Rightarrow\widehat{ODN}=\widehat{ADB}+\widehat{ADN}=115^0\\ MH\text{//}AN\Rightarrow\widehat{MHA}=\widehat{HAN}\\ \Rightarrow\widehat{MHB}+\widehat{MBH}=\widehat{BAD}+\widehat{NAD}\\ \Rightarrow\widehat{MHB}=\widehat{NAD}\\ \Rightarrow\Delta MHB\sim\Delta AND\left(g.g\right)\\ \Rightarrow\dfrac{MB}{AD}=\dfrac{HB}{ND}\Rightarrow MB\cdot NC=AD\cdot HB\left(1\right)\)

\(\left\{{}\begin{matrix}\widehat{OHB}=\widehat{AOD}=90^0\\\widehat{HBO}=\widehat{ODA}\end{matrix}\right.\Rightarrow\Delta HBO\sim\Delta ODA\\ \Rightarrow\dfrac{HB}{OD}=\dfrac{OB}{AD}\Rightarrow HB\cdot AD=OB\cdot OD\left(2\right)\\ \left(1\right)\left(2\right)\Rightarrow\dfrac{MB}{OD}=\dfrac{OB}{ND}\\ \text{Mà }\widehat{MBO}=\widehat{NDO}\\ \Rightarrow\Delta MBO\sim\Delta ODN\left(c.g.c\right)\\ \Rightarrow\widehat{MOB}=\widehat{OND}\Rightarrow\widehat{MOB}+\widehat{NOD}=\widehat{OND}+\widehat{NOD}\\ \Rightarrow\widehat{MOB}+\widehat{NOD}=180^0-\widehat{NDO}=65^0\\ \Rightarrow180^0-\widehat{MON}=65^0\\ \Rightarrow\widehat{MON}=115^0\)

Để 2 đt cắt tại 1 điểm trên Oy

\(\Leftrightarrow\left\{{}\begin{matrix}-2x-m+3=0\\\dfrac{1}{2}x+5-3m=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=3-m\\2x+10-6m=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x=3-m\\2x=6m-10\end{matrix}\right.\\ \Leftrightarrow3-m=6m-10\Leftrightarrow m=\dfrac{13}{7}\)

\(\Rightarrow3x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x-2y}{5-6}=\dfrac{-4}{-1}=4\\ \Rightarrow\left\{{}\begin{matrix}x=20\\y=12\end{matrix}\right.\)