Cậu viết đề rõ hơn bằng công thức nha
(\(\dfrac{2+x}{2-x}+\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\)) :\(\dfrac{x^2-3x}{2x^2-x^3}\)
\(\)
a) ĐKXĐ :
* 2 - x # 0 => x # 2
* x2 - 4 => x # 2 ; x # -2
* 2 + x # 0 => x # -2
* x2( 2 - x) # 0 => x # 0 ; x # 2
* x( x - 3) # 0 => x # 0 ; x # 3
Vậy,...
Rút gọn mk k chép lại đề nha
P = \(\dfrac{-\left(x+2\right)^2+4x^2+\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}:\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
P = \(\dfrac{-x^2-4x-4+4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}.\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
P = \(\dfrac{4x^2-8x}{\left(x-2\right)\left(x+2\right)}.\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
P = \(\dfrac{4x\left(x-2\right)}{x+2}.\dfrac{-x^2}{x\left(x-3\right)}=\dfrac{-4x^2\left(x-2\right)}{\left(x+2\right)\left(x-3\right)}\)
Mk nghĩ là đề cần xem lại nhé , lẻ lắm
