(x+1)²+(y-2)¹⁰=0

Vì \(\left\{{}\begin{matrix}\left(x+1\right)^2≥0∀x\\\left(y-2\right)^{10}≥0∀y\end{matrix}\right.\)\(\Rightarrow\left(x+1\right)^2+\ge\left(y-2\right)^{10}0∀x,y\)

Dáu "=" xảy ra <=> \(\left[{}\begin{matrix}x+1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

Vì 5a27b chia 5 dư 1 \(\Rightarrow\left[{}\begin{matrix}b=1\\b=6\end{matrix}\right.\) ( do b là chữ số )

Vì 5a27b ⋮ 3 => 5+a+2+7+b ⋮ 3 => 14 + a + b ⋮ 3 mà \(\left[{}\begin{matrix}b=1\\b=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}14+a+1⋮3\\14+a+6⋮3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}15+a⋮3\\20+a⋮3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a∈\left\{0;3;6;9\right\}\\a∈\left\{1;4;7\right\}\end{matrix}\right.\)

Ta có :

(x-20).(x-8).(x-2022) = 0

\(\Rightarrow\left[{}\begin{matrix}x-20=0\\x-8=0\\x-2022=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=20\\x=8\\x=2022\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=20\\x=8\\x=2022\end{matrix}\right.\)