\(a,ĐK:\dfrac{3x-2}{5}\ge0\Leftrightarrow3x-2\ge0\left(5>0\right)\Leftrightarrow x\ge\dfrac{2}{3}\\ b,ĐK:\dfrac{2x-3}{-3}\ge0\Leftrightarrow2x-3\le0\left(-3< 0\right)\Leftrightarrow x\le\dfrac{3}{2}\)

\(\sqrt{x+7}\) có nghĩa \(\Leftrightarrow x+7\ge0\Leftrightarrow x\ge-7\)

\(\sqrt{x-5}\) có nghĩa \(\Leftrightarrow x-5\ge0\Leftrightarrow x\ge5\)

\(\sqrt{3-\dfrac{2}{3}x}\) có nghĩa \(\Leftrightarrow3-\dfrac{2}{3}x\ge0\Leftrightarrow-\dfrac{2}{3}x\ge-3\Leftrightarrow x\le\dfrac{9}{2}\)

\(\sqrt{5-3x}\) có nghĩa \(\Leftrightarrow5-3x\ge0\Leftrightarrow-3x\ge-5\Leftrightarrow x\le\dfrac{5}{3}\)

Bài 1:

$x-1=|2x-1|\geq 0\Rightarrow x\geq 1$

$\Rightarrow 2x-1>0\Rightarrow |2x-1|=2x-1$. Khi đó:

$2x-1=x-1\Leftrightarrow x=0$  (không thỏa mãn vì $x\geq 1$)

Vậy không tồn tại $x$ thỏa đề.

Bài 2:

Nếu $x\geq \frac{1}{3}$ thì:

$3x-1=2x+3$

$\Leftrightarrow x=4$ (tm)

Nếu $x< \frac{1}{3}$ thì:

$1-3x=2x+3$

$\Leftrightarrow -2=5x\Leftrightarrow x=\frac{-2}{5}$ (tm)

Vậy......

Bài 3: Xét các TH sau:

TH1: $x\geq 2$ thì:

$x-1+x-2=3$

$2x-3=3$

$2x=6$

$x=3$ (thỏa mãn)

TH2: $1\leq x< 2$ thì:

$x-1+2-x=3$

$1=3$ (vô lý- loại)

TH3: $x< 1$

$1-x+2-x=3$

$3-2x=3$

$2x=0$

$x=0$ (thỏa mãn)

Bài 1 : 

\(\left|2x-1\right|=x-1\)ĐK : \(x\ge1\)

TH1 : \(2x-1=x-1\Leftrightarrow x=0\)(ktm)

TH2 : \(2x-1=1-x\Leftrightarrow3x=2\Leftrightarrow x=-\frac{2}{3}\)(ktm)

Vậy biểu thức ko có x thỏa mãn 

Bài 2 : 

\(\left|3x-1\right|=2x+3\)ĐK : x >= -3/2 

TH1 : \(3x-1=2x+3\Leftrightarrow x=4\)

TH2 : \(3x-1=-2x-3\Leftrightarrow5x=-2\Leftrightarrow x=-\frac{2}{5}\)

\(2,\\ a,\sqrt{4x-4}+\sqrt{9x-9}-\sqrt{25x-25}=7\left(x\ge1\right)\\ \Leftrightarrow2\sqrt{x-1}+3\sqrt{x-1}-5\sqrt{x-1}=7\\ \Leftrightarrow0\sqrt{x-1}=7\Leftrightarrow x\in\varnothing\\ b,\sqrt{2x^2-3}=4\left(x\le-\dfrac{\sqrt{6}}{2};\dfrac{\sqrt{6}}{2}\le x\right)\\ \Leftrightarrow2x^2-3=16\\ \Leftrightarrow x^2=\dfrac{19}{2}\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{19}{2}}\left(tm\right)\\x=-\sqrt{\dfrac{19}{2}}\left(tm\right)\end{matrix}\right.\)

\(1,\\ A=\sqrt{5+4x}+\sqrt{7-3x}\\ ĐKXĐ:\left\{{}\begin{matrix}5+4x\ge0\\7-3x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{5}{4}\\x\le\dfrac{7}{3}\end{matrix}\right.\)

Bài 2:

a) \(\sqrt{4x-4}+\sqrt{9x-9}-\sqrt{25x-25}=7\left(đk:x\ge1\right)\)

\(\Leftrightarrow2\sqrt{x-1}+3\sqrt{x-2}-5\sqrt{x-1}=7\)

\(\Leftrightarrow0=7\left(VLý\right)\)

Vậy \(S=\varnothing\)

b) \(\sqrt{2x^2-3}=4\left(đk:-\sqrt{\dfrac{3}{2}}\ge x\ge\sqrt{\dfrac{3}{2}}\right)\)

\(\Leftrightarrow2x^2-3=16\)

\(\Leftrightarrow2x^2=19\Leftrightarrow x^2=\dfrac{19}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{19}{2}}\left(tm\right)\\x=-\sqrt{\dfrac{19}{2}}\left(tm\right)\end{matrix}\right.\)

\(\dfrac{3x}{7}\ge0\)

\(\Leftrightarrow3x\ge0\)

\(\Leftrightarrow x\ge0\)

Để biểu thức \(\sqrt{\dfrac{3x}{7}}\) có nghĩa thì 3x/7 ≥0

⇒x≥0

ĐKXĐ: \(x\ge0\)

- 3 x + 4 có nghĩa khi -3x +4 ≥ 0

-3x + 4 ≥ 0 ⇔ -3x ≥ -4

⇔ x ≤ 4 3

\(-3x+4\ge0\\ \Rightarrow-3x\ge-4\\ \Rightarrow x\le\dfrac{4}{3}\)