Question

giải phương trình: \(3+\sqrt{3+\sqrt{x}}=x\)

Answer 1

\(x\ge3\)

\(\sqrt{3+\sqrt{x}}=\left(x-3\right)\Leftrightarrow3+\sqrt{x}=x^2-6x+9\Leftrightarrow x^2-6x-\sqrt{x}+6=0\)

\(\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)-6\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\)

\(\left(\sqrt{x}-1\right)\left(x\sqrt{x}+x-5\sqrt{x}-6\right)=0\) x >/ 3

\(\left(x\sqrt{x}+x-5\sqrt{x}-6\right)=0\)

\(x\left(\sqrt{x}+2\right)-\sqrt{x}\left(\sqrt{x}+2\right)-3\left(\sqrt{x}-2\right)=0\)

\(\left(\sqrt{x}+2\right)\left(x-\sqrt{x}-3\right)=0\)

\(\left(x-\sqrt{x}-3\right)=0\)

giải tiếp nhé.

 

Answer 2

em moi co lop 7 thoi