thung nho: 7,9+4,5=12,4(l)

thung to:12,4+4,3=16,7(l)

\(3\sqrt{8}-4\sqrt{2}+5\sqrt{32}\\ =3\sqrt{4.2}-4\sqrt{2}+5\sqrt{16.2}\\ =3.2\sqrt{2}-4\sqrt{2}+5.4\sqrt{2}\\ =6\sqrt{2}-4\sqrt{2}+20\sqrt{2}\\ =22\sqrt{2}\)

\(x-\sqrt{x+1}-1=0\\ =>\sqrt{x+1}=x-1\\ < =>\left\{{}\begin{matrix}x+1=\left(x-1\right)^2\\x-1\ge0\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}x+1=x^2-2x+1\\x\ge1\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}x^2-3x=0\\x\ge1\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}x\left(x-3\right)=0\\x\ge1\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\\x\ge1\end{matrix}\right.< =>x=3\)

Vậy \(S=\left\{3\right\}\)

c) ĐKXĐ : \(\left\{{}\begin{matrix}x+3\ge0\\9x^2-x-4\ge0\end{matrix}\right.\)

Ta có \(2\sqrt{x+3}=9x^2-x-4\)

<=> \(9x^2-\left(x+2\sqrt{x+3}+4\right)=0\)

<=> \(9x^2-\left(\sqrt{x+3}+1\right)^2=0\)

<=> \(\left(3x-\sqrt{x+3}-1\right).\left(3x+\sqrt{x+3}+1\right)=0\)

<=> \(\left[{}\begin{matrix}3x-1=\sqrt{x+3}\left(1\right)\\3x+1=-\sqrt{x+3}\left(2\right)\end{matrix}\right.\)

Giải (2) ta có \(\left\{{}\begin{matrix}\left(3x+1\right)^2=x+3\\3x+1\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x^2+5x-2=0\\x\le-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-5\pm97}{18}\\x\le-\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{-5-\sqrt{97}}{18}\)(tm ĐKXĐ)

Giải (1) ta có \(\left\{{}\begin{matrix}\left(3x-1\right)^2=x+3\\3x-1\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x^2-7x-2=0\\x\ge\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{9}\end{matrix}\right.\\x\ge\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow x=1\) (tm ĐKXĐ) 

Vậy tập nghiệm phương trình : S = \(\left\{1;\dfrac{-5-\sqrt{97}}{18}\right\}\)