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

\(\Leftrightarrow\dfrac{3\left(x-3\right)+4\left(2x-1\right)+2x}{12}=0\)

\(\Leftrightarrow3x-9+8x-4+2x=0\)

\(\Leftrightarrow13x-13=0\)

\(\Leftrightarrow13x=13\)

\(\Leftrightarrow x=1\)

\(b,\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\-x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-6\end{matrix}\right.\)

a, x-3/4+2x-1/3=-x/6

b,(x-3)(2x-1)=(2x-1)(2x+3)

c,6/x-1-4/x-3+8/x^2-4x+3=0

a 2x-3=4x+6
(=) 2x-4x=6+3
(=)-2x=9
(=)x=-\(\dfrac{9}{2}\)

c: =>x(x-1+x+3)=0

=>x(2x+2)=0

=>x=-1 hoặc x=0

d: =>x(x+1)-x(x-3)=4x

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

=>4x=4x(luôn đúng)

a)ĐKXĐ: \(x\notin\left\{0;-1\right\}\)

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

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

Suy ra: \(x^2-1+x-2x+1=0\)

\(\Leftrightarrow x^2-x=0\)

\(\Leftrightarrow x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)

Vậy: S={1}

b) ĐKXĐ: \(x\notin\left\{3;-3\right\}\)

Ta có: \(\dfrac{5}{x-3}-\dfrac{2x-3}{x+3}=\dfrac{2x\left(1-x\right)}{x^2-9}\)

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

Suy ra: \(5x+15-2x^2+6x+3x-9-2x+2x^2=0\)

\(\Leftrightarrow12x+6=0\)

\(\Leftrightarrow12x=-6\)

hay \(x=-\dfrac{1}{2}\)(thỏa ĐK)

Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)

đk: \(_{x+1\ne0\Leftrightarrow x\ne-1}\)\(\dfrac{1-x}{x+1}+3=\dfrac{2x-3}{x+1}\Leftrightarrow\dfrac{1-x}{x+1}+\dfrac{3\left(x+1\right)}{x+1}=\dfrac{2x+3}{x-1}\Leftrightarrow1-x+3x+3-2x-3=0\Leftrightarrow-2x+1=0\Leftrightarrow-2x=-1\Leftrightarrow x=0,5\)

f/ \(3xy\left(x+y\right)-\left(x+y\right)\left(x^2+y^2+2xy\right)+y^3=27\)

\(3x^2y+3xy^2-\left(x+y\right)\left(x+y\right)^2+y^3=27\)

\(3x^2y+3xy^3-\left(x+y\right)^3+y^3=27\)

\(3x^2y+3xy^3-\left(x^3+3x^2y+3xy^2+b^3\right)+y^3=27\)

\(-x^3=27\)

\(x=-3\)

Bài 1:

a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(6x-9+4-2x=-3\)

\(4x=-2\)

\(x=-\frac{1}{2}\)

b/ \(2x\left(x^2-2\right)+x^2\left(1-2x\right)-x^2=-12\)

\(2x^3-4x+x^2-2x^3-x^2=-12\)

\(-4x=-12\)

\(x=\frac{1}{3}\)

Bài 2:

a/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(6x^2+9x-6x^2-15x+4x+10=8\)

\(-2x=8\)

\(x=-4\)

b/ \(4x\left(x-1\right)-3\left(x^2-5\right)-x^2=\left(x-3\right)-\left(x+4\right)\)

\(4x^2-4x-3x^2+15-x^2=-7\)

\(-4x=-22\)

\(x=\frac{11}{2}\)

c/ \(2\left(3x-1\right)\left(2x+5\right)-6\left(2x-1\right)\left(x+2\right)=-6\)

\(6x-2\left(2x+5\right)-12x+6\left(x+2\right)=-6\)

\(6x-4x-10-12x+6x+12=-6\)

\(-4x=-8\)

\(x=2\)