\(c2:3x+5x^2\ge-6+5x+5x^2\Leftrightarrow2x-6\le0\Leftrightarrow x\le3\)
\(c3:-x+7=6a-1\Leftrightarrow x=-\left(6a-1-7\right)=8-6a>0\Leftrightarrow a< \dfrac{4}{3}\)
\(c4:pt\Leftrightarrow\left(2019-x\right)^3+\left(2021-x\right)^3+\left[2x-4040\right]^3=0\left(1\right)\)
\(đặt:\left[\left(2019-x\right);\left(2021-x\right)\right]=\left\{u;v\right\}\)
\(\Rightarrow2x-4040=x-2019+x-2021=-u-v\)
\(\left(1\right)\Leftrightarrow u^3+v^3+\left(-u-v\right)^3=0\Leftrightarrow-3uv\left(u+v\right)=0\Leftrightarrow\left[{}\begin{matrix}u.v=0\\u=-v\end{matrix}\right.\)
\(u.v=0\Leftrightarrow\left[{}\begin{matrix}u=0\Leftrightarrow2019-x=0\Leftrightarrow x=2019\\v=0\Leftrightarrow2021-x=0\Leftrightarrow x=2021\end{matrix}\right.\)
\(u=-v\Leftrightarrow2019-x=x-2021\Leftrightarrow x=2020\)
\(\Rightarrow S=\left\{2019;2020;2021\right\}\)
Câu 2 :
\(\Leftrightarrow3x+5x^2+6-5x-5x^2\ge0\Leftrightarrow-2x+6\ge0\Leftrightarrow x\le3\)
Câu 4 :
PT <=> \(2019-x+2021-x+2x-4040=0\Leftrightarrow2019+2021-4040=0\)
( đúng )
Vậy pt có vô số nghiệm
