Question

Answer 1

Answer 2

ĐKXĐ: \(x\ge0\)

\(x+6\sqrt{x}+9-x+3=0\)

\(\Leftrightarrow6\sqrt{x}+12=0\)

Do \(6\sqrt{x}\ge0;\forall x\Rightarrow6\sqrt{x}+12>0\)

\(\Rightarrow\)Phương trình vô nghiệm

Answer 3

Ta có: \(\left(\sqrt{x}+3\right)^2-x+3=0\)

\(\Leftrightarrow x+6\sqrt{x}+9-x+3=0\)

\(\Leftrightarrow6\sqrt{x}=-12\left(vôlý\right)\)