a: =>|x-3/2|=2

\(\Leftrightarrow x-\dfrac{3}{2}\in\left\{2;-2\right\}\)

hay \(x\in\left\{\dfrac{7}{2};-\dfrac{1}{2}\right\}\)

f: \(\Leftrightarrow\left[{}\begin{matrix}2x+3=x-2\\2x+3=2-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{3}\end{matrix}\right.\)

Lời giải:
ĐKXĐ: $x\neq 0; \frac{-3}{2}; \frac{-1}{2}; -3$

PT $\Leftrightarrow (\frac{1}{x}-\frac{3}{2x+1})+(\frac{5}{2x+3}-\frac{4}{x+3})=0$

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

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

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

TH1: $1-x=0\Leftrightarrow x=1$ (tm) 

TH2: $\frac{1}{x(2x+1)}+\frac{3}{(2x+3)(x+3)}=0$

$\Rightarrow (2x+3)(x+3)+3x(2x+1)=0$

$\Leftrightarrow 8x^2+12x+9=0$

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

$\Rightarrow (2x+3)^2=x^2=0$ (vô lý) 

Do đó $x=1$ là nghiệm duy nhất.

a: \(\Leftrightarrow\dfrac{y+5}{y\left(y-5\right)}-\dfrac{y-5}{2y\left(y+5\right)}=\dfrac{y+25}{2\left(y-5\right)\left(y+5\right)}\)

\(\Leftrightarrow2\left(y+5\right)^2-\left(y-5\right)^2=y^2+25y\)

=>\(2y^2+20y+50-y^2+10y-25=y^2+25y\)

=>30y+25=25y

=>5y=-25

=>y=-5(loại)

b: \(\Leftrightarrow x\left(x+1\right)+x\left(x-3\right)=4x\)

=>x^2+x+x^2-3x-4x=0

=>2x^2-6x=0

=>2x(x-3)=0

=>x=0(nhận) hoặc x=3(loại)

c: =>x^2-9-6(2x+7)=-13(x+3)

=>x^2-9-12x-42+13x+39=0

=>x^2+x-6=0

=>(x+3)(x-2)=0

=>x=2(nhận) hoặc x=-3(loại)

\(\left(x-1\right)^3-\left(x-1\right)\left(2x-3\right)\left(3x-5\right)+\left(2x-3\right)^3-\left(x-1\right)\left(2x-3\right)\left(3x-5\right)+\left(3x-5\right)^3-\left(x-1\right)\left(2x-3\right)\left(3x-5\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(\left(x-1\right)^2-\left(2x-3\right)\left(3x-5\right)\right)+\left(2x-3\right)\left(\left(2x-3\right)^2-\left(x-1\right)\left(3x-5\right)\right)+\left(3x-5\right)\left(\left(3x-5\right)^2-\left(x-1\right)\left(2x-3\right)\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(7-5x\right)+\left(2x-3\right)\left(x-2\right)^2+\left(3x-5\right)\left(x-2\right)\left(7x-11\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(\left(x-1\right)\left(7-5x\right)+\left(2x-3\right)\left(x-2\right)+\left(3x-5\right)\left(7x-11\right)\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(18x^2-63x+54\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\18x^2-63x+54=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{2}\end{matrix}\right.\)

`(3x-1)/(x-1)-(2x+5)/(x+3)+4/(x^2+2x-3)=1(x ne 1,-3)`

`<=>((3x-1)(x+3))/(x^2+2x-3)-((2x+5)(x-1))/(x^2+2x-3)+4/(x^2+2x-3)=(x^2+2x-3)/(x^2+2x-3)`

`<=>(3x-1)(x+3)-(2x+5)(x-1)+4=x^2+2x-3`

`<=>3x^2+8x-3-2x^2-3x+5+4=x^2+2x-3`

`<=>x^2+5x+6=x^2+2x-3`

`<=>3x=-9`

`<=>x=-3(loại)`

Vậy `S={cancel0}`

ĐKXĐ: \(x\notin\left\{1;-3\right\}\)

Ta có: \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}+\dfrac{4}{x^2+2x-3}=1\)

\(\Leftrightarrow\dfrac{\left(3x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(2x+5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}+\dfrac{4}{\left(x+3\right)\left(x-1\right)}=\dfrac{x^2+2x-3}{\left(x+3\right)\left(x-1\right)}\)

\(\Leftrightarrow\dfrac{3x^2+9x-x-3-\left(2x^2-2x+5x-5\right)+4}{\left(x+3\right)\left(x-1\right)}=\dfrac{x^2+2x-3}{\left(x+3\right)\left(x-1\right)}\)

\(\Leftrightarrow\dfrac{3x^2+8x-3-\left(2x^2+3x-5\right)+4}{\left(x+3\right)\left(x-1\right)}=\dfrac{x^2+2x-3}{\left(x+3\right)\left(x-1\right)}\)

\(\Leftrightarrow\dfrac{3x^2+8x+1-2x^2-3x+5}{\left(x+3\right)\left(x-1\right)}=\dfrac{x^2+2x-3}{\left(x+3\right)\left(x-1\right)}\)

Suy ra: \(x^2+5x+6-x^2-2x+3=0\)

\(\Leftrightarrow3x+9=0\)

\(\Leftrightarrow3x=-9\)

hay x=-3(Không nhận)

Vậy: \(S=\varnothing\)

Không ạ

a, ĐKXĐ:\(x\ne-5\)

\(\dfrac{2x-5}{x+5}=3\\ \Rightarrow2x-5=3\left(x+5\right)\\ \Leftrightarrow3x+15-2x+5=0\\ \Leftrightarrow x+20=0\\ \Leftrightarrow x=-20\)

b, ĐKXĐ:\(x\ne3\)

\(\dfrac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}=0\\ \Rightarrow x^2+2x-3x-6=0\\ \Leftrightarrow x^2-x-6=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(3x+6\right)=0\\ \Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\left(tm\right)\\x=3\left(ktm\right)\end{matrix}\right.\)

c, ĐKXĐ:\(\left\{{}\begin{matrix}x\ne-1\\x\ne3\end{matrix}\right.\)

\(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\\ \Leftrightarrow x\left(\dfrac{1}{2\left(x-3\right)}+\dfrac{1}{2\left(x+1\right)}-\dfrac{2}{\left(x+1\right)\left(x-3\right)}\right)=0\\ \Leftrightarrow x\left(\dfrac{x+1}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x-3}{2\left(x+1\right)\left(x-3\right)}-\dfrac{4}{2\left(x+1\right)\left(x-3\right)}\right)=0\\ \Leftrightarrow x.\dfrac{x+1+x-3-4}{2\left(x-3\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x\left(2x-6\right)}{2\left(x-3\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{2x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x}{x+1}=0\\ \Rightarrow x=0\left(tm\right)\)

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

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

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)