a) \(\dfrac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}\)
= \(\dfrac{\left(x^3+x^2\right)-\left(4x+4\right)}{\left(x^3+x^2\right)+\left(7x^2+7x\right)+\left(10x+10\right)}\)
=\(\dfrac{x^2\left(x+1\right)-4\left(x+1\right)}{x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)}\)
= \(\dfrac{\left(x^2-4\right)\left(x+1\right)}{\left(x^2+7x+10\right)\left(x+1\right)}\)
= \(\dfrac{\left(x+2\right)\left(x-2\right)\left(x+1\right)}{\left[\left(x^2+2x\right)+\left(5x+10\right)\right]\left(x+1\right)}\)
= \(\dfrac{\left(x+2\right)\left(x-2\right)}{x\left(x+2\right)+5\left(x+2\right)}\)
= \(\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+5\right)\left(x+2\right)}\)
= \(\dfrac{x-2}{x+5}\)
b) \(\dfrac{x^3+3x^2-4}{x^3-3x+2}\)
= \(\dfrac{x^3-x^2+4x^2+4x-4x-4}{x^3-x-2x+2}\)
= \(\dfrac{\left(x^3-x^2\right)+\left(4x^2-4x\right)+\left(4x-4\right)}{\left(x^3-x\right)-\left(2x-2\right)}\)
= \(\dfrac{x^2\left(x-1\right)+4x\left(x-1\right)+4\left(x-1\right)}{x^2\left(x-1\right)-2\left(x-1\right)}\)
= \(\dfrac{\left(x^2+4x+4\right)\left(x-1\right)}{\left(x^2-2\right)\left(x-1\right)}\)
= \(\dfrac{\left(x+2\right)^2\left(x-1\right)}{\left(x^2-2\right)\left(x-1\right)}\)
= \(\dfrac{\left(x+2\right)^2}{x^2-2}\)
Nhớ tik nha...
chữa lại câu b nhé:
b) \(\dfrac{x^3+3x^2-4}{x^3-3x+2}\)
= \(\dfrac{x^3-x^2+4x^2+4x-4x-4}{x^3-x-2x+2}\)
=\(\dfrac{\left(x^3-x^2\right)+\left(4x^2-4x\right)+\left(4x-4\right)}{\left(x^3-x\right)-\left(2x-2\right)}\)
= \(\dfrac{x^2\left(x-1\right)+4x\left(x-1\right)+4\left(x-1\right)}{x\left(x^2-1\right)-2\left(x-1\right)}\)
= \(\dfrac{\left(x^2+4x+4\right)\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)}\)
= \(\dfrac{\left(x+2\right)^2\left(x-1\right)}{\left(x-1\right)\left[x\left(x+1\right)-2\right]}\)
= \(\dfrac{\left(x+2\right)^2}{x^2+x-2}\)
Đúng chưa bn...