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

1) \(\sqrt[]{3x+7}-5< 0\)

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

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

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

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

\(\left\{{}\begin{matrix}\dfrac{6}{3x-2}-2\sqrt{1-y}=1\\\dfrac{2}{3x-2}+\sqrt{1-y}=2\end{matrix}\right.\) (x \(\ne\) \(\dfrac{2}{3}\); y \(\le\) 1)

Đặt \(\dfrac{1}{3x-2}=a\); \(\sqrt{1-y}=b\) 

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}6a-2b=1\\2a+b=2\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}6a-2b=1\\4a+2b=4\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}10a=5\\4a+2b=4\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=\dfrac{1}{2}\\4\cdot\dfrac{1}{2}+2b=4\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\) 

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}\dfrac{1}{3x-2}=2\\\sqrt{1-y}=1\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}2\left(3x-2\right)=1\\1-y=1\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=\dfrac{5}{6}\\y=0\end{matrix}\right.\) (TM)

Vậy ...

Chúc bn học tốt!

ĐK:x khác 2/3, y<_1

đặt 1/3x-2=u,căn (1-y) Ta có hệ

6u-2v=1

2u+v=2

sau đó giải hệ và trả ẩn b tự lm nha

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............

Đặt \(\sqrt{3x+1}=a\)

\(\Rightarrow\frac{a^2-1}{\sqrt{a^2+9}}=a-1\)

\(\Leftrightarrow\left(a-1\right)\left(\frac{a+1}{\sqrt{a^2+9}}-1\right)=0\)