Question

[NGUYỄN ĐỖ BẢO VY]

P=\(1+\dfrac{x+3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)
Rút gọn P

Answer 1

[Vương Thúy Phương]

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

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

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

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

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

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

   = 1 + \(\dfrac{x-2}{6}\)

  = \(\dfrac{6+x-2}{6}\)

= \(\dfrac{x+4}{6}\)