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

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

\(ĐKXĐ:x\ne1\)

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

\(\Rightarrow4x^2-3x+5-\left(1+2x\right)\left(x-1\right)-6\left(x^2+x+1\right)\)

\(\Rightarrow4x^2-3x+5-\left(x-1+2x^2-2x\right)-6x^2-6x-6\)

\(\Rightarrow4x^2-3x+5-x+1-2x^2+2x-6x^2-6x-6\)

\(\Rightarrow-4x^2-8x\)

⇒-4x(x-4)

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

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

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

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

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

b) Ta có: \(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)

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

\(=\dfrac{9x^2+3x+2x-6x^2}{\left(1-3x\right)\left(1+3x\right)}:\dfrac{2x\left(3x+5\right)}{\left(1-3x\right)^2}\)

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

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

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

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

\(=\left(\dfrac{9}{x\left(x-3\right)\left(x+3\right)}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x\left(x+3\right)}-\dfrac{x}{3\left(x+3\right)}\right)\)

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

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

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

\(=\dfrac{-3}{x-3}\)

\(\dfrac{x^2-50}{3x^2-9x}\div\dfrac{2x^2+10x}{x^2-9}\)

\(\Leftrightarrow\dfrac{x^2-50}{3x\left(x-3\right)}\div\dfrac{2x^2+10x}{\left(x-3\right)\left(x+3\right)}\)

MTC: 3x(x-3)(x+3)

\(\dfrac{(x^2-50)\left(x+3\right)}{3x\left(x-3\right)\left(x+3\right)}\div\dfrac{3x(2x^2+10x)}{3x\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow\)(x²-50)(x+3):3x(2x²+10x)

\(\Rightarrow\)(x³+3x²-50x-150):6x³+30x²

a) Ta có: \(\dfrac{x^2-50}{3x^2-9x}:\dfrac{2x^2+10x}{x^2-9}\)

\(=\dfrac{x^2-50}{3x\left(x-3\right)}\cdot\dfrac{\left(x-3\right)\left(x+3\right)}{2x\left(x+5\right)}\)

\(=\dfrac{\left(x^2-50\right)\left(x+3\right)}{6x^2\left(x+5\right)}\)

b) Ta có: \(\dfrac{-3x^2}{2x+1}:\dfrac{-9}{4x^2-1}\)

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

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

a: =2x^5-15x^3-x^2-2x^5-x^3=-16x^3-x^2

b: =x^3+3x^2-2x-3x^2-9x+6

=x^3-11x+6

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

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

a) \(6x^3\left(\dfrac{1}{3}x^2-\dfrac{5}{2}-\dfrac{1}{6}\right)-2x^5-x^3\)

\(=6x^3\left(\dfrac{1}{3}x^2-\dfrac{16}{6}\right)-2x^5-x^3\)

\(=2x^5-16x^3-2x^5-x^3\)

\(=-17x^3\)

b) \(\left(x+3\right)\left(x^2+3x-2\right)\)

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

\(=x^3+6x^2+7x-6\)

c) \(\left(4x^3-4x^2-5x+4\right):\left(2x+1\right)\)

\(=2x^2+4x^3-2x-4x^2-\dfrac{5}{2}-5x+\dfrac{2}{x}+4\)

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

`a)3x(2x^2-3x+4)`

`=6x^3-9x^2+12x`

______________________________________________

`b)(x+3)^2+(3x-2)(x+4)`

`=x^2+6x+9+3x^2+12x-2x-8`

`=4x^2+16x+1`

______________________________________________

`c)[2x-4]/[x-1]+[2x+2]/[x^2-1]`       `ĐK: x \ne +-1`

`=[(2x-4)(x+1)+2x+2]/[(x-1)(x+1)]`

`=[2x^2+2x-4x-4+2x+2]/[(x-1)(x+1)]`

`=[2x^2-2]/[x^2-1]`

`=2`

hếp

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

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

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

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

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