Question

Cho biểu thức:

A = (\(\dfrac{x^2}{x^3-4x}\) + \(\dfrac{6}{6-3x}\) + \(\dfrac{1}{x+2}\)) : (x - 2 + \(\dfrac{10-x^2}{x+2}\) )

Tìm tập xác định.

Với giá trị nào của x, giá trị của biểu thức rút gọn bằng 2??

Answer 1

\(A=\left(\dfrac{x^2}{x^3-4x}+\dfrac{6}{6-3x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)ĐK : \(x\ne-2;2\)

\(=\left(\dfrac{x}{x-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\left(\dfrac{x^2-4+10-x^2}{x+2}\right)\)

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

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

\(=\dfrac{x^3-4x+x^2-2x+24}{\left(x-4\right)\left(x-2\right)\left(x+2\right)}:\dfrac{6}{x+2}=\dfrac{x^3+x^2-6x+24}{\left(x-4\right)\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{6}\)

\(=\dfrac{x^3+x^2-6x+24}{6\left(x-4\right)\left(x-2\right)}=\dfrac{\left(x+4\right)\left(x^2-3x+6\right)}{6\left(x-4\right)\left(x-2\right)}\)

Answer 2

P/s : mình thấy đề này cứ sai sai ở đâu ý ! 

b, Ta có : \(\dfrac{\left(x+4\right)\left(x^2-3x+6\right)}{6\left(x-4\right)\left(x-2\right)}=2\)

\(\Leftrightarrow\dfrac{\left(x+4\right)\left(x^2-3x+6\right)-12\left(x-4\right)\left(x-2\right)}{6\left(x-4\right)\left(x-2\right)}=0\)

\(\Rightarrow x^3-11x^2+66x-72=0\)