Lời giải:
ĐKXĐ: $x\geq \frac{-1}{3}$

PT $\Leftrightarrow \frac{x}{\sqrt{x+2}}=\sqrt{3x+1}-\sqrt{x+1}$

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

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

Xét các TH:

TH1: $x=0$ (thỏa mãn)

TH2: $\frac{1}{\sqrt{x+2}}-\frac{2}{\sqrt{3x+1}+\sqrt{x+1}}$

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

$\Rightarrow 4x+2+2\sqrt{(3x+1)(x+1)}=4(x+2)$

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

$\Rightarrow (3x+1)(x+1)=9$

$\Leftrightarrow 3x^2+4x-8=0$

$\Rightarrow x=\frac{-2\pm 2\sqrt{7}}{3}$

Kết hợp với ĐKXĐ suy ra $x=\frac{-2+2\sqrt{7}}{3}$

Vậy............

1) \(\sqrt[]{3x+7}-5< 0\)

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

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

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

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

a) \(\dfrac{3x^2+1}{\sqrt{x-1}}=\dfrac{4}{\sqrt{x-1}}\)

ĐKXĐ: \(x>1\)

\(3x^2+1=4\)

\(3x^2=3\)

\(x^2=1\)

\(x=\pm1\)

=> Pt vô nghiệm

b) ĐKXĐ: x>-4

\(x^2+3x+4=x+4\)

\(x^2+2x=0\)

\(x\left(x+2\right)=0\)

\(\left[{}\begin{matrix}x=0\\x+2=0\Leftrightarrow x=-2\end{matrix}\right.\)

c) Đkxđ: \(3x-2>0\Leftrightarrow x>\dfrac{2}{3}\)
Pt \(\Leftrightarrow3x^2-x-2=3x-2\)
\(\Leftrightarrow3x^2-4x=0\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0\left(l\right)\\x=\dfrac{4}{3}\left(tm\right)\end{matrix}\right.\)
Vậy \(x=\dfrac{4}{3}\) là nghiệm của phương trình.
d) Đkxđ: \(x\ne1\)
\(2x+3+\dfrac{4}{x-1}=\dfrac{x^2+3}{x-1}\) \(\Leftrightarrow\dfrac{\left(2x+3\right)\left(x-1\right)}{x-1}+\dfrac{4}{x-1}=\dfrac{x^2+3}{x-1}\)
\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)+4=x^2+3\)
\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(l\right)\\x=-2\left(tm\right)\end{matrix}\right.\)
Vậy \(x=-2\) là nghiệm của phương trình.

ĐK: \(x\ge2\)

\(\dfrac{\sqrt{x^2+1}-\sqrt{x+1}}{x^2+\sqrt{3x-6}}\ge0\)

\(\Leftrightarrow\sqrt{x^2+1}-\sqrt{x+1}\ge0\)

\(\Leftrightarrow\sqrt{x^2+1}\ge\sqrt{x+1}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\x^2+1\ge x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2-x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-1\le x\le0\\x\ge1\end{matrix}\right.\)

Kết hợp điều kiện xác định ta được \(x\ge2\)

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 $x=\frac{1-\sqrt{17}}{24}$

Cách 2:

ĐKXĐ:...........

PT $\Leftrightarrow 12x^2-3x-1=4x\sqrt{3x+1}$

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

Đặt $4x=a; \sqrt{3x+1}=b$ thì pt trở thành:

$\frac{3}{4}a^2-b^2=ab$

$\Leftrightarrow 3a^2-4b^2-4ab=0$

$\Leftrightarrow (a-2b)(3a+2b)=0$

Nếu $a-2b=0\Leftrightarrow 4x=2\sqrt{3x+1}$

$\Rightarrow 4x^2=3x+1$ và $x\geq 0$

$\Rightarrow x=1$ (chọn) hoặc $x=-\frac{1}{4}$ (loại do $x\geq 0$)

Nếu $3a+2b=0$

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

Bình phương lên ta cũng thu được $x=\frac{1-\sqrt{17}}{24}$

a.

ĐKXĐ: \(x>0\)

\(\sqrt{x\left(x+3\right)}+2\sqrt{x+2}=2x+\sqrt{\dfrac{\left(x+2\right)\left(x+3\right)}{x}}\)

\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}-\sqrt{x+3}\right)+\sqrt{\dfrac{x+2}{x}}\left(\sqrt{x+3}-2\sqrt{x}\right)=0\)

\(\Leftrightarrow\sqrt{x}\left(\dfrac{4x-x-3}{2\sqrt{x}+\sqrt{x+3}}\right)-\sqrt{\dfrac{x+2}{x}}\left(\dfrac{4x-x-3}{\sqrt{x+3}+2\sqrt{x}}\right)=0\)

\(\Leftrightarrow\dfrac{3\left(x-1\right)}{2\sqrt{x}+\sqrt{x+3}}\left(\sqrt{x}-\sqrt{\dfrac{x+2}{x}}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{x+2}{x}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-1\left(loại\right)\end{matrix}\right.\)

b.

ĐKXĐ: \(x\ge-\dfrac{1}{2};x\ne1-\sqrt{2}\)

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

\(\Leftrightarrow\sqrt{x+2}\left(\sqrt{2x+1}-\sqrt{x+2}\right)-x\left(\sqrt{2x+1}-\sqrt{x+2}\right)=0\)

\(\Leftrightarrow\left(\sqrt{2x+1}-\sqrt{x+2}\right)\left(\sqrt{x+2}-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+1}=\sqrt{x+2}\\\sqrt{x+2}=x\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=x+2\\x^2-x-2=0\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-1\left(loại\right)\end{matrix}\right.\)

\(ĐK:-1\le x< 0;x\ge1\\ PT\Leftrightarrow x+2\sqrt{x-\dfrac{1}{x}}=3+\dfrac{1}{x}\\ \Leftrightarrow x-\dfrac{1}{x}+2\sqrt{x-\dfrac{1}{x}}-3=0\)

Đặt \(\sqrt{x-\dfrac{1}{x}}=a\ge0\)

\(PT\Leftrightarrow a^2+2a-3=0\\ \Leftrightarrow\left(a-1\right)\left(a+3\right)=0\\ \Leftrightarrow a=1\left(a\ge0\right)\\ \Leftrightarrow x-\dfrac{1}{x}=1\\ \Leftrightarrow x^2-x-1=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{5}}{2}\left(tm\right)\\x=\dfrac{1+\sqrt{5}}{2}\left(tm\right)\end{matrix}\right.\)