Hummmm

1) \(\sqrt[]{3x+7}-5< 0\)

\(\Leftrightarrow\sqrt[]{3x+7}< 5\)

\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)

\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)

\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)

Đk:\(x\ne0;x\ge-\dfrac{1}{3}\)

Pt \(\Leftrightarrow12x^2-3x-1=4x\sqrt{3x+1}\)

\(\Leftrightarrow16x^2=4x^2+4x\sqrt{3x+1}+3x+1\)

\(\Leftrightarrow16x^2=\left(2x+\sqrt{3x+1}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=2x+\sqrt{3x+1}\\4x=-\left(2x+\sqrt{3x+1}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\sqrt{3x+1}\left(1\right)\\6x=-\sqrt{3x+1}\left(2\right)\end{matrix}\right.\)

TH1 \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\4x^2=3x+1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(x-1\right)\left(4x+1\right)=0\end{matrix}\right.\)\(\Rightarrow x=1\) (thỏa)

TH2\(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\36x^2=3x+1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\\left[{}\begin{matrix}x=\dfrac{1+\sqrt{17}}{24}\\x=\dfrac{1-\sqrt{17}}{24}\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow x=\dfrac{1-\sqrt{17}}{24}\)(tm)

Vậy...

Lời giải:
ĐKXĐ: $x\ge \frac{-1}{3}; x\neq 0$

PT \(\Leftrightarrow 3(x-1)+\frac{x-1}{4x}=\sqrt{3x+1}-2\)

\(\Leftrightarrow 3(x-1)+\frac{x-1}{4x}=\frac{3(x-1)}{\sqrt{3x+1}+2}\)

\(\Leftrightarrow (x-1)(3+\frac{1}{4x}-\frac{3}{\sqrt{3x+1}+2})=0\)

Nếu $x-1=0\Leftrightarrow x=1$ (tm)

Nếu $3+\frac{1}{4x}-\frac{3}{\sqrt{3x+1}+2}=0$

$\Leftrightarrow 12x\sqrt{3x+1}+12x+\sqrt{3x+1}+2=0$

$\Leftrightarrow \sqrt{3x+1}(12x+1)=-(12x+2)$

Từ đây suy ra $x\leq \frac{-1}{6}$

Bình phương 2 vế:

$(3x+1)(12x+1)^2=[(12x+1)+1]^2$

$\Leftrightarrow 3x(12x+1)^2=2(12x+1)+1$

$\Leftrightarrow 144x^3+24x^2-7x-1=0$

$\Leftrightarrow (4x+1)(36x^2-3x-1)=0$

Vì $x\leq \frac{-1}{6}$ nên $x=\frac{1-\sqrt{17}}{24}$