Question

Rút gọn P :

Làm hộ mình ạ :)) 3 bạn lam nhanh và đúng đầu tiên mình sẽ qua wall tick đúng 20 câu trả lời ( 20 likes)...Hứa ạ :))

Mình chưa bùng bao giờ bùng đâu ạ :))

Càng nhanh thì mình càng tặng tick :))

Answer 1

Ta có :

P = ( \(\frac{2+x}{2-x}+\frac{4x^2}{x-4}-\frac{2-x}{2+x}\)): \(\frac{x^2-3}{2x^2-x^3}\)

=>P = \(\left(\frac{2+x}{2-x}-\frac{4x^2}{4-x^2}-\frac{2-x}{2+x}\right).\frac{2x^2-x^3}{x^2-3x}\)

=>P = \(\left(\frac{\left(2+x\right)\left(2+x\right)}{\left(2-x\right)\left(2+x\right)}-\frac{4x^2}{\left(2-x\right)\left(2+x\right)}-\frac{\left(2-x\right)\left(2-x\right)}{\left(2+x\right)\left(2-x\right)}\right).\frac{x^2\left(2-x\right)}{x\left(x-3\right)}\)

=>P = \(\frac{\left(2+x\right)^2-4x^2-\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}.\frac{x\left(2-x\right)}{x-3}\)

=>P = \(\frac{\left(2+x-2+x\right)\left(2+x+2-x\right)-4x^2}{\left(2-x\right)\left(2+x\right)}.\frac{x\left(2-x\right)}{x-3}\)

=>P = \(\frac{4x-4x^2}{\left(2-x\right)\left(2+x\right)}.\frac{x\left(2-x\right)}{x-3}\)

=> P = \(\frac{4x^2\left(1-x\right)}{\left(2+x\right)\left(x-3\right)}\)

Answer 2

P= \(\left(\frac{2+x}{2-x}+\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\frac{x^2-3x}{2x^2-x^3} \)
=\(\left[\frac{\left(2+x\right)\left(2+x\right)}{\left(2-x\right)\left(2+x\right)}+\frac{-4x^2}{\left(2-x\right)\left(2+x\right)}-\frac{\left(2-x\right)\left(2-x\right)}{\left(2-x\right)\left(2+x\right)}\right]:\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
= \(\left[\frac{4+4x+x^2}{\left(2-x\right)\left(2+x\right)}+\frac{-4x^2}{\left(2-x\right)\left(2+x\right)}-\frac{4-4x+x^2}{\left(2-x\right)\left(2+x\right)}\right].\frac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
= \(\left[\frac{4+4x+x^2-4x^2-4+4x-x^2}{\left(2-x\right)\left(2+x\right)}\right].\frac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
= \(\frac{8x-4x^2}{\left(2-x\right)\left(2+x\right)}.\frac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
= \(\frac{4x\left(2-x\right)}{\left(2-x\right)\left(2+x\right)}.\frac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
= \(\frac{4\left(2-x\right)}{2+x}.\frac{x^2}{x-3}\)
= \(\frac{8x^2-4x^3}{\left(2+x\right)\left(x-3\right)}=\frac{8x^2-4x^3}{x^2-x-6}\)
= \(\frac{-4x^2\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}=\frac{-4x^2}{x+3}\)