a:Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

=>3x-9-10x+2=-4

=>-7x-7=-4

=>-7x=3

=>x=-3/7

b: =>\(\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)

=>\(2\left(5-x\right)+7\left(x-2\right)=4\left(x-1\right)+x\)

=>10-2x+7x-14=4x-4+x

=>5x-4=5x-4

=>0x=0(luôn đúng)

Vậy: S=R\{0;2}

\(\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}=\frac{3}{4}\)

\(\frac{1}{x-1}-\frac{1}{x+2}=\frac{3}{4}\)

tự tính nhé bạn

+ Pt thứ nhất :

Ta có mẫu thức chung là : \(2\left(x-3\right)\left(x+1\right)\)

\(\Rightarrow\left[{}\begin{matrix}x\ne2\\x-3\ne0\\x+1\ne0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x\ne2\\x\ne3\\x\ne-1\end{matrix}\right.\)

Vậy \(ĐKXĐ\) là :\(x\ne2;3;-1\)

+ Pt thứ hai : 

Ta có mẫu thức chung là : \(\left(x-2\right)\left(x+3\right)\)

\(\Rightarrow\left[{}\begin{matrix}x-2\ne0\\x+3\ne0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x\ne2\\x\ne-3\end{matrix}\right.\)

Vậy \(DKXD:\) \(\) \(x\ne2;-3\)

làm ơn giúp 🙏🙏🙏

a: =>1/3x+2/5x-2/5=0

=>11/15x-2/5=0

=>11/15x=2/5

=>x=2/5:11/15=2/5*15/11=30/55=6/11

b: =>-5x-1-1/2x+1/3=x

=>-11/2x-2/3-x=0

=>-13/2x=2/3

=>x=-2/3:13/2=-2/3*2/13=-4/39

c: (x+1/2)(2/3-2x)=0

=>x+1/2=0 hoặc 2/3-2x=0

=>x=1/3 hoặc x=-1/2

d: 9(3x+1)^2=16

=>(3x+1)^2=16/9

=>3x+1=4/3 hoặc 3x+1=-4/3

=>3x=1/3 hoặc 3x=-7/3

=>x=1/9 hoặc x=-7/9

\(1,\) thiếu đề

\(2,\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)

\(\Leftrightarrow\dfrac{5\left(5x+2\right)}{30}-\dfrac{10\left(8x-1\right)}{30}=\dfrac{6\left(4x+2\right)}{30}-\dfrac{150}{30}\)

\(\Leftrightarrow5\left(5x+2\right)-10\left(8x-1\right)=6\left(4x+2\right)-150\)

\(\Leftrightarrow25x+10-80x+10=24x+12-150\)

\(\Leftrightarrow-55x+20=24x-138\)

\(\Leftrightarrow24x-138+55x-20=0\)

\(\Leftrightarrow79x-158=0\)

\(\Leftrightarrow x=2\)

\(3,ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne-1\\x\ne3\end{matrix}\right.\\ \dfrac{x}{2x-6}+\dfrac{x}{2x-2}=\dfrac{-2x}{\left(x+1\right)\left(3-x\right)}\)

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

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

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

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

\(\Leftrightarrow x\left(\dfrac{x^2-1}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}+\dfrac{x^2-2x-3}{2\left(x-1\right)\left(x-3\right)\left(x+1\right)}-\dfrac{4x-4}{2\left(x+1\right)\left(x-3\right)\left(x-1\right)}\right)=0\)

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

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

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

\(\Leftrightarrow x.\dfrac{x}{\left(x-1\right)\left(x+1\right)}=0\)

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

\(\Leftrightarrow x=0\)

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

=>-x^2+2x-1=10x-5x^2-11x-22

=>-x^2+2x-1=-5x^2-x-22

=>4x^2+3x+21=0

=>PTVN

b: \(\Leftrightarrow\left(x+10\right)\left(x+4\right)+3\left(x+4\right)\left(x-2\right)=4\left(x+10\right)\left(x-2\right)\)

=>x^2+14x+40+3(x^2+2x-8)=4(x^2+8x-20)

=>x^2+14x+40+3x^2+6x-24=4x^2+32x-80

=>20x+16=32x-80

=>-12x=-96

=>x=8

c: \(\Leftrightarrow6\left(x-3\right)+7\left(x-5\right)=13x+4\)

=>6x-18+7x-35=13x+4

=>-53=4(loại)

d: =>3(2x-1)-5(x-2)=3(x+7)

=>6x-3-5x+10=3x+21

=>3x+21=x+7

=>x=-7

e: =>x^3-6x^2+12x-8-x^3-3x^2-3x-1=-9x^2+1

=>-9x^2+9x-9=-9x^2+1

=>9x=10

=>x=10/9

ĐKXĐ: \(x\notin\left\{-1;-2;-3;-4\right\}\)

Ta có: \(\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+4}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{x+4}{\left(x+1\right)\left(x+4\right)}-\dfrac{x+1}{\left(x+1\right)\left(x+4\right)}=\dfrac{1}{6}\)

\(\Leftrightarrow\dfrac{x+4-x-1}{\left(x+1\right)\left(x+4\right)}=\dfrac{x^2+5x+4}{6\left(x+1\right)\left(x+4\right)}\)

\(\Leftrightarrow\dfrac{18}{6\left(x+1\right)\left(x+4\right)}=\dfrac{x^2+5x+4}{6\left(x+1\right)\left(x+4\right)}\)

Suy ra: \(x^2+5x+4=18\)

\(\Leftrightarrow x^2+5x-14=0\)

\(\Leftrightarrow x^2+7x-2x-14=0\)

\(\Leftrightarrow x\left(x+7\right)-2\left(x+7\right)=0\)

\(\Leftrightarrow\left(x+7\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

Vậy: S={-7;2}

ĐKXĐ: $x \neq -1;-2;-3;-4$

$pt⇔\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}=\dfrac{1}{6}$

$⇔\dfrac{1}{x+1}-\dfrac{1}{x+4}=\dfrac{1}{6}$

$⇔\dfrac{3}{(x+1)(x+4)}=\dfrac{1}{6}$

$⇔x^2+5x+4=18$

$⇔x^2+5x-14=0$

$⇔(x-2)(x+7)=0$

$⇔$ \(\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\)(t/m)

Vậy...