a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)

\(=\dfrac{3x+2-3x+2-3x+6}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)

b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)

\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)

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

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

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

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

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

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

\(a,=\dfrac{4x+8}{x^2+2x}=\dfrac{4\left(x+2\right)}{x\left(x+2\right)}=\dfrac{4}{x}\\ b,=\dfrac{\left(2x-3\right)-\left(2x-4\right)}{x-2}=\dfrac{2x-3-2x+4}{x-2}=\dfrac{1}{x-2}\\ c,=\dfrac{2x-1-3x-2}{x+3}=\dfrac{-x-3}{x+3}=\dfrac{-\left(x+3\right)}{x+3}=-1\\ d,=\dfrac{11x-18+x}{2x-3}=\dfrac{12x-18}{2x-3}=\dfrac{6\left(2x-3\right)}{2x-3}=6\)

\(e,=\dfrac{3x-6-9x+3}{2x+1}=\dfrac{-6x-3}{2x+1}=\dfrac{-3\left(2x+1\right)}{2x+1}=-3\)

a: \(=\dfrac{2x-2x+y}{2\left(2x-y\right)}=\dfrac{y}{2\left(2x-y\right)}\)

b: \(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{2\left(x-1\right)}\)

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

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

c: \(=\dfrac{1}{x+2}+\dfrac{x+8}{3x\left(x+2\right)}\)

\(=\dfrac{3x+x+8}{3x\left(x+2\right)}=\dfrac{4x+8}{3x\left(x+2\right)}=\dfrac{4}{3x}\)

d: \(=\dfrac{4x+6-2x^2+3x+2x+1}{\left(2x-3\right)\left(2x+3\right)}\)

\(=\dfrac{-2x^2+9x+7}{\left(2x-3\right)\left(2x+3\right)}\)

a) \(\dfrac{x^3+2x}{x^3+1}+\dfrac{2x}{x^2-x+1}+\dfrac{1}{x+1}\)

\(=\dfrac{x^3+2x}{\left(x+1\right)\left(x^2-x+1\right)}+\dfrac{2x}{x^2-x+1}+\dfrac{1}{x+1}\) MTC: \(\left(x+1\right)\left(x^2-x+1\right)\)

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

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

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

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

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

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

b) \(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{3x-2}{2x-4x^2}\)

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

\(=\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}-\dfrac{3x-2}{2x\left(2x-1\right)}\) MTC: \(2x\left(2x-1\right)\)

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

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

\(=\dfrac{\left(2x-1-6x^2+3x\right)+\left(6x^2-4x\right)-\left(3x-2\right)}{2x\left(2x-1\right)}\)

\(=\dfrac{2x-1-6x^2+3x+6x^2-4x-3x+2}{2x\left(2x-1\right)}\)

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

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

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

b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)

\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)

\(=\dfrac{2y^2+8y+12}{y-1}\)

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

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

\(=\dfrac{6x^2+3x+2x-4x^2}{\left(1-2x\right)\left(1+2x\right)}:\dfrac{2x^2+5x}{\left(1-2x\right)^2}\)

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

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

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

Đặt `A=(1-3x)/(2x)+(3x-2)/(2x-1)+(3x-2)/(2x-4x^2)`

`=(2x(3x-2))/(2x(2x-1))-((3x-1)(2x-1))/(2x(2x-1))-(3x-2)/(2x(2x-1))`

`=(6x^2-4x-6x^2+5x-1-3x+2)/(2x(2x-1))`

`=(-2x+1)/(2x(2x-1))`

`=-1/(2x)`

`2x=1/(483)`

`=>A=-1/(1/483)=-483`

\(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{3x-2}{2x-4x^2}\)

⇔\(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{3x-2}{2x\left(1-2x\right)}\)

⇔\(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}-\dfrac{(3x-2)}{2x\left(2x-1\right)}\)

ĐKXĐ: \(\left\{{}\begin{matrix}2x\ne0\\2x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\2x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne\dfrac{1}{2}\end{matrix}\right.\)

MTC: 2x(2x-1)

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

\(\Rightarrow1-3x\left(2x-1\right)+\left(3x-2\right)\left(2x\right)-\left(3x-2\right)\)

\(\Leftrightarrow1-6x^2+3x+6x^2-4x-3x+2\\ \Rightarrow-4x+3\)

Thay \(x=\dfrac{1}{966}\)vào biểu thức trên

ta có -4x+3= \(-4\times\dfrac{1}{966}+3=\dfrac{-4}{966}+3=\dfrac{-2}{483}+3=\dfrac{-2}{483}+\dfrac{1449}{483}=\dfrac{1447}{483}\)

a) Ta có: \(\dfrac{x^2+38x+4}{2x^2+17x+1}-\dfrac{3x^2-4x-2}{2x^2+17x+1}\)

\(=\dfrac{x^2+38x+4-3x^2+4x+2}{2x^2+17x+1}\)

\(=\dfrac{-2x^2+42x+6}{2x^2+17x+1}\)

c) Ta có: \(C=\dfrac{-x}{3x-2}+\dfrac{7x-4}{3x-2}\)

\(=\dfrac{-x+7x-4}{3x-2}\)

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

a) Ta có: \(\dfrac{x}{x-1}-\dfrac{2}{x-1}\)

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

b) Ta có: \(\dfrac{4+4x}{3x^2+6x}+\dfrac{x}{3x+6}\)

\(=\dfrac{4+4x}{x\left(3x+6\right)}+\dfrac{x^2}{x\left(3x+6\right)}\)

\(=\dfrac{x^2+4x+4}{3x\left(x+2\right)}\)

\(=\dfrac{\left(x+2\right)^2}{3x\left(x+2\right)}\)

\(=\dfrac{x+2}{3x}\)

c) Ta có: \(\dfrac{x^2-2x}{x-1}\cdot\dfrac{1}{x}:\dfrac{x^2-4}{x^2-2x+1}\)

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

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

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