a, đk x ≥ 1; x ≥ 2
\(\sqrt{\left(\sqrt{x}-1\right)^2}+\sqrt{\left(\sqrt{x}+2\right)^2}=3\\ \sqrt{x}-1+\sqrt{x}+2=3\\ 2\sqrt{x}+1=3\\ 2\sqrt{x}=2\\ \sqrt{x}=1\\ x=1\left(thoaman\right)\\ \)
b,
\(đkx\ge1\sqrt{x-1+2\sqrt{x-1}+1}=2\\ \sqrt{\left(\sqrt{x-1}+1\right)^2}=2\\ \sqrt{x-1}+1=2\\ \sqrt{x-1}=1\\ x-1=1\\ x=2\left(thoaman\right)\)