Ta có: \(\sqrt{2x+7}-6=x\)
\(\Leftrightarrow\sqrt{2x+7}=x+6\)
\(\Leftrightarrow x^2+12x+36-2x-7=0\)
\(\Leftrightarrow x^2+10x+29=0\)(Vô lý)
Vậy: \(S=\varnothing\)
Điều kiện : x ≥ 0
\(\sqrt{2x+97}-6=x\text{⇔}\sqrt{2x+97}=x+6\\ \text{⇔}2x+97=x^2+12x+36\text{⇔}x^2+10x-61=0\\ \text{⇔}\left[{}\begin{matrix}x=-5+\sqrt{86}\\x=-5-\sqrt{86}\end{matrix}\right.\)
\(\sqrt{2x+97}-6=x\)
\(\Leftrightarrow\sqrt{2x+97}=x+6\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+6\ge0\\2x+97=\left(x+6\right)^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-6\\x^2+10x-61=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-6\\\left[{}\begin{matrix}x=-5+\sqrt{86}\\x=-5-\sqrt{86}\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow x=-5+\sqrt{86}\)
Vậy..,
