Question

1.

2. 

Answer 1

\(\Leftrightarrow x^2-6x+17=8\)

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

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

Answer 2

\(PT\Leftrightarrow x^2-6x+17=8\Leftrightarrow x^2-6x+9=0\\ \Leftrightarrow x=3\)

Vậy PT có 1 nghiệm nguyên

\(B=\sqrt{9x+\sqrt[3]{\left(6\sqrt{x}+1\right)^3}}=\sqrt{9x+6\sqrt{x}+1}=\sqrt{\left(3\sqrt{x}+1\right)^2}=3\sqrt{x}+1\)