Question

Answer 1

a, đk x ≥ 2; x ≥ 3

\(\sqrt{\left(x-2\right)^2}+\sqrt{\left(x-3\right)^2}=1\\ x-2+x-3=1\\ 2x-5=1\\ 2x=6\\ x=3\left(thoaman\right)\)

b, đk x ≥ 0

\(\sqrt{\left(\sqrt{x}-2\right)^2}+\sqrt{\left(\sqrt{x}-3\right)^2}=1\\ \sqrt{x}-2+\sqrt{x}-3=1\\ 2\sqrt{x}-5=1\\ 2\sqrt{x}=6\\ \sqrt{x}=3\\ x=9\left(thoaman\right)\)