Hummmm

1) \(\sqrt[]{3x+7}-5< 0\)

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

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

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

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

1: ĐKXĐ: x>=8/3

\(\sqrt{3x-8}-\sqrt{x+1}=\frac{2x-11}{5}\)

=>\(\sqrt{3x-8}-1+2-\sqrt{x+1}=\frac{2x-11}{5}+1\)

=>\(\frac{3x-8-1}{\sqrt{3x-8}+1}+\frac{4-x-1}{2+\sqrt{x+1}}=\frac{2x-11+5}{5}\)

=>\(\left(x-3\right)\left(\frac{3}{\sqrt{3x-8}+1}-\frac{1}{2+\sqrt{x+1}}-\frac25\right)=0\)

=>x-3=0

=>x=3(nhận)

3: ĐKXĐ: -5/2<=x<=5/2

Đặt \(a=\sqrt{5+2x};b=\sqrt{5-2x}\)

=>\(ab=\sqrt{\left(5+2x\right)\left(5-2x\right)}=\sqrt{25-4x^2}\)

Theo đề, ta có: a+b+5=3ab

=>3ab-a-b-5=0

=>a(3b-1)-b+1/3-16/3=0

=>\(3a\left(b-\frac13\right)-\left(b-\frac13\right)=\frac{16}{3}\)

=>\(\left(b-\frac13\right)\left(3a-1\right)=\frac{16}{3}\)

=>(3a-1)(3b-1)=16

=>(3a-1;3b-1)∈{(1;16);(16;1);(2;8);(8;2);(4;4)}

=>(3a;3b)∈{(2;17);(17;2);(3;9);(9;3);(5;5)}

=>(a;b)∈{(2/3;17/3);(17/3;2/3);(1;3);(3;1);(5/3;5/3)}

mà a<>b

nên (a;b)∈{(2/3;17/3);(17/3;2/3);(1;3);(3;1)}

TH1: a=2/3 và b=17/3

=>\(\begin{cases}5+2x=\frac49\\ 5-2x=\frac{289}{9}\end{cases}\Rightarrow\begin{cases}2x=\frac49-5=\frac49-\frac{45}{9}=-\frac{41}{9}\\ 2x=5-\frac{289}{9}=-\frac{244}{9}\end{cases}\)

=>x∈∅

TH2: a=17/3 và b=2/3

=>\(\begin{cases}5+2x=\frac{289}{9}\\ 5-2x=\frac49\end{cases}\Rightarrow\begin{cases}2x=\frac{289}{9}-5=\frac{244}{9}\\ 2x=5-\frac49=\frac{41}{9}\end{cases}\)

=>x∈∅

TH3: a=1 và b=3

=>5+2x=1 và 5-2x=9

=>2x=-4 và 2x=5-9=-4

=>x=-2(nhận)

TH4: a=3 và b=1

=>5+2x=9 và 5-2x=1

=>2x=4 và 2x=4

=>x=2(nhận)

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