Question

Giải phương trình: \(\sqrt{1+\left(x+2\right)\sqrt{1+\left(x+3\right)\left(x+5\right)}}=2023x+1\)

Answer 1

ĐKXĐ : \(x\ge-2\)

\(\sqrt{1+\left(x+2\right).\sqrt{1+\left(x+3\right).\left(x+5\right)}}=2023x+1\)

\(\Leftrightarrow\sqrt{1+\left(x+2\right).\sqrt{x^2+8x+16}}=2023x+1\)

\(\Leftrightarrow\sqrt{1+\left(x+2\right).\left(x+4\right)}=2023x+1\) (Do \(x\ge-2\Rightarrow x+4>0\))

\(\Leftrightarrow\sqrt{x^2+6x+9}=2023x+1\)

\(\Leftrightarrow x+3=2023x+1\) (Do \(x\ge-2\Rightarrow x+3>0\)

\(\Leftrightarrow x=\dfrac{1}{1011}\)(tm) 

Vậy tập nghiệm \(S=\left\{\dfrac{1}{1011}\right\}\)