Question

Answer 1

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

Answer 2

Bạn lấy nhân tử chung là x ở tử chia cho x ở dưới mẫu nốt hộ mình nhé