\(a\text{) }x^2+\sqrt{x+2019}=2019\left(x\ge-2019\right)\\ \Leftrightarrow x^2+x+\frac{1}{4}=\left(x+2019\right)-\sqrt{x+2019}+\frac{1}{4}\\ \Leftrightarrow\left(x+\frac{1}{2}\right)^2=\left(\sqrt{x+2019}-\frac{1}{2}\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\sqrt{x+2019}-\frac{1}{2}\left(1\right)\\x+\frac{1}{2}=\frac{1}{2}-\sqrt{x+2019}\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x+\frac{1}{4}=\sqrt{x+2019}\\ ĐK:x\ge-\frac{1}{4}\\ \Leftrightarrow\left(x+\frac{1}{4}\right)^2=x+2019\\ \Leftrightarrow\left(x+\frac{1}{4}\right)^2=x+2019\\ \Leftrightarrow x^2+\frac{1}{2}x+\frac{1}{16}-x-2019=0\\ \Leftrightarrow x^2-\frac{1}{2}x+\frac{1}{16}-2019=0\\ \Leftrightarrow\left(x-\frac{1}{4}\right)^2-2019=0\\ \Leftrightarrow\left(x-\frac{1}{4}-\sqrt{2019}\right)\left(x-\frac{1}{4}+\sqrt{2019}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{4}-\sqrt{2019}=0\\x-\frac{1}{4}+\sqrt{2019}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{4\sqrt{2019}+1}{4}\left(T/m\right)\\x=\frac{-\sqrt{2019}+1}{4}\left(K^o\text{ }T/m\right)\end{matrix}\right.\)
\(\left(2\right)\Leftrightarrow x=-\sqrt{x+2019}\\ ĐK:-2019\le x\le0\\ \Leftrightarrow x^2=x+2019\\ \Leftrightarrow x^2-x-2019=0\\ \Leftrightarrow x^2-x-2019=0\\ \Leftrightarrow.....\)
\(b\text{) }x+\sqrt{2-x^2}+x\sqrt{2-x^2}=3\)
\(Đặt\text{ }\sqrt{2-x^2}=y\)
\(\Rightarrow\left\{{}\begin{matrix}x+y+xy=3\\x^2+y^2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+2y+2xy=6\\2x^2+2y^2=4\end{matrix}\right.\\\Leftrightarrow2x^2+2y^2-\left(2x+2y+2xy\right)=-2\\ \Leftrightarrow\left(x^2-2x+1\right)+\left(y^2-2y+1\right)+\left(x^2-2xy+y^2\right)=0\\ \Leftrightarrow\left(x-1\right)^2+\left(y-1\right)^2+\left(x-y\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-1=0\\y-1=0\\x-y=0\end{matrix}\right.\Leftrightarrow x=1\)
Vậy....