\(b,\Delta=\left(m+5\right)^2-4\left(-m+6\right)\ge0\Leftrightarrow\left[{}\begin{matrix}m\le-7-4\sqrt{3}\\m\ge-7+4\sqrt{3}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x1+x2=m+5\\2x1+3x2=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x1+2x2=2m+10\\2x1+3x2=13\end{matrix}\right.\)\(\)

\(\Rightarrow x2=13-2m-10=3-2m\Rightarrow x1=m+5-x2=m+5-3+2m=3m+2\)

\(x1x2=6-m\Rightarrow\left(3-2m\right)\left(3m+2\right)=6-m\Leftrightarrow\left[{}\begin{matrix}m=0\left(tm\right)\\m=1\left(tm\right)\end{matrix}\right.\)

\(c,\Delta'=\left(m+1\right)^2-\left(m^2-2m+29\right)\ge0\Leftrightarrow m\ge7\)

\(\Rightarrow\left\{{}\begin{matrix}x1+x2=2m+2\\x1=2x2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x2=\dfrac{2m+2}{3}\\x1=\dfrac{2\left(2m+2\right)}{3}\end{matrix}\right.\)

\(\Rightarrow x1.x2=\dfrac{\left(2m+2\right).2\left(2m+2\right)}{9}=m^2-2m+29\Leftrightarrow\left[{}\begin{matrix}m=11\left(tm\right)\\m=23\left(tm\right)\end{matrix}\right.\)

\(c2:3x+5x^2\ge-6+5x+5x^2\Leftrightarrow2x-6\le0\Leftrightarrow x\le3\)

\(c3:-x+7=6a-1\Leftrightarrow x=-\left(6a-1-7\right)=8-6a>0\Leftrightarrow a< \dfrac{4}{3}\)

\(c4:pt\Leftrightarrow\left(2019-x\right)^3+\left(2021-x\right)^3+\left[2x-4040\right]^3=0\left(1\right)\)

\(đặt:\left[\left(2019-x\right);\left(2021-x\right)\right]=\left\{u;v\right\}\)

\(\Rightarrow2x-4040=x-2019+x-2021=-u-v\)

\(\left(1\right)\Leftrightarrow u^3+v^3+\left(-u-v\right)^3=0\Leftrightarrow-3uv\left(u+v\right)=0\Leftrightarrow\left[{}\begin{matrix}u.v=0\\u=-v\end{matrix}\right.\)

\(u.v=0\Leftrightarrow\left[{}\begin{matrix}u=0\Leftrightarrow2019-x=0\Leftrightarrow x=2019\\v=0\Leftrightarrow2021-x=0\Leftrightarrow x=2021\end{matrix}\right.\)

\(u=-v\Leftrightarrow2019-x=x-2021\Leftrightarrow x=2020\)

\(\Rightarrow S=\left\{2019;2020;2021\right\}\)

có đoạn  bạn sửa lại tí nhé tại lúc đầu mình đọc đề thành \(a+b+c=2022\)

\(M\ge a+b+c-\left(\dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{2}\right)\ge a+b+c-\dfrac{a+b+c}{2}=\dfrac{a+b+c}{2}\ge\dfrac{2022}{2}=1011\)

\(\dfrac{a^3}{a^2+bc}=a-\dfrac{abc}{a^2+bc}\ge a-\dfrac{abc}{2a\sqrt{bc}}=a-\dfrac{\sqrt{bc}}{2}\)

\(\dfrac{b^3}{b^2+ca}\ge b-\dfrac{\sqrt{ac}}{2};\dfrac{c^3}{c^2+ab}\ge c-\dfrac{\sqrt{ab}}{2}\)

\(\Rightarrow M\ge a+b+c-\left(\dfrac{\sqrt{ab}}{2}+\dfrac{\sqrt{bc}}{2}+\dfrac{\sqrt{ca}}{2}\right)=2022-\left(\dfrac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}\right)\)

\(do:\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le a+b+c\)

\(\Rightarrow M\ge2022-\dfrac{a+b+c}{2}=2022-\dfrac{2022}{2}=1011\)

\(min_M=2021\Leftrightarrow a=b=c=674\)

\(\)\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\Rightarrow2\ge\dfrac{4}{x+y}\Leftrightarrow x+y\ge2\)(chắc bài cho x,y>0?

\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\Rightarrow\sqrt{xy}\ge1\Leftrightarrow xy\ge1\)

\(D=\dfrac{1}{x+2y}+\dfrac{1}{2x+y}=\dfrac{1}{x+y+y}+\dfrac{1}{x+x+y}\le\dfrac{1}{y+2}+\dfrac{1}{x+2}\)

\(cm:\dfrac{1}{x+2}+\dfrac{1}{y+2}\le\dfrac{2}{3}\Leftrightarrow\dfrac{x+y+4}{\left(x+2\right)\left(y+2\right)}\le\dfrac{2}{3}\)

\(\Leftrightarrow2\left(x+2\right)\left(y+2\right)\ge3\left(x+y+4\right)\Leftrightarrow4x+4y+8+2xy\ge3x+3y+12\Leftrightarrow x+y+2xy\ge4\left(1\right)\)

\(x+y\ge2;xy\ge1\Rightarrow\left(1\right)đúng\Rightarrow D\le\dfrac{2}{3}\Rightarrow dấu"="xayra\Leftrightarrow x=y=1\)

\(\left(x\ne-y;x>\dfrac{y}{2}\right)\Rightarrow\left\{{}\begin{matrix}\dfrac{4}{\sqrt{2x-y}}-\dfrac{21}{x+y}=\dfrac{1}{2}\\\dfrac{3}{\sqrt{2x-y}}+\dfrac{7-\left(x+y\right)}{x+y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{\sqrt{2x-y}}-\dfrac{21}{x+y}=\dfrac{1}{2}\\\dfrac{3}{\sqrt{2x-y}}+\dfrac{7}{x+y}=2\end{matrix}\right.\)

\(đặt:\dfrac{1}{\sqrt{2x-y}}=a>0;\dfrac{1}{x+y}=b\)

\(\Rightarrow\left\{{}\begin{matrix}4a-21b=\dfrac{1}{2}\\3a+7b=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\left(tm\right)\\b=\dfrac{1}{14}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{2x-y}}=\dfrac{1}{2}\\\dfrac{1}{x+y}=\dfrac{1}{14}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=4\\x+y=14\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=8\end{matrix}\right.\)(thỏa)

\(\left(dk:x\ne-\dfrac{2}{3};x\ne-1\right)pt\Leftrightarrow\dfrac{2x}{3x^2-x+2}-\dfrac{7x-3x^2-5x-2}{3x^2+5x+2}=0\Leftrightarrow\dfrac{2x}{3x^2-x+2}-\dfrac{3x^2+12x+2}{3x^2+5x+2}=0\left(1\right)\)

\(x=0\) \(không\) \(là\) \(nghiệm\left(1\right)\)

\(x\ne0\Rightarrow\left(1\right)\Leftrightarrow\dfrac{2}{3x-1+\dfrac{2}{x}}-\dfrac{3x+12+\dfrac{2}{x}}{3x+5+\dfrac{2}{x}}=0\)

\(đặt:3x+\dfrac{2}{x}=t\) \(do:x\ne-\dfrac{2}{3};x\ne-1;\Rightarrow t\ne-5\)

\(x>0\Rightarrow t\ge2\sqrt{3.2}=2\sqrt{6}\)

\(x< 0\Rightarrow-t\ge2\sqrt{6}\Rightarrow t\le-2\sqrt{6}\Rightarrow\left[{}\begin{matrix}t\ne-5;t\le-2\sqrt{6}\\t\ge2\sqrt{6}\end{matrix}\right.\)

\(\Rightarrow\dfrac{2}{t-1}-\dfrac{t+12}{t+5}=0\Rightarrow2\left(t+5\right)-\left(t+12\right)\left(t-1\right)=0\Leftrightarrow\left[{}\begin{matrix}t=-11\left(tm\right)\\t=2\left(ktm\right)\end{matrix}\right.\)

\(t=-11=3x+\dfrac{2}{x}\Leftrightarrow3x^2+2=-11x\Leftrightarrow3x^2+11x+2=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11+\sqrt{97}}{6}\left(tm\right)\\x=\dfrac{-11-\sqrt{97}}{6}\left(tm\right)\end{matrix}\right.\)

\(a,\left\{{}\begin{matrix}mx-y=2m\\x-my=m+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m^2x-my=2m^2\\x-my=m+1\end{matrix}\right.\)

\(\Leftrightarrow m^2x-x=2m^2-m-1\Leftrightarrow x\left(m^2-1\right)=2m^2-m-1\)

\(ycầuđềbài\Leftrightarrow m^2-1\ne0\Leftrightarrow m\ne\pm-1\)

\(b,\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2m^2-m-1}{m^2-1}=\dfrac{\left(m-1\right)\left(2m+1\right)}{m^2-1}=\dfrac{2m+1}{m+1}=2+\dfrac{-2}{m+1}\\y=mx-2m=\dfrac{m\left(2m+1\right)-2m^2-2m}{m+1}=\dfrac{-m}{m+1}=-1+\dfrac{1}{m+1}\end{matrix}\right.\)

\(\left(x;y\right)\in Z\Leftrightarrow\left\{{}\begin{matrix}m\ne\pm1\\m+1\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\\m+1\inƯ\left(1\right)=\left\{1;-1\right\}\end{matrix}\right.\)

\(\Rightarrow m=0;m=-2\)