Câu hỏi của Phạm Phương Anh

Question

Rút gọn phân thức sau

B=$\dfrac{x^3+2x^2y-xy^2-2y^3}{x^2+3xy+2y^2}$

nguyễn ngọc dinh · Accepted Answer

$B=\dfrac{x^3+2x^2y-xy^2-2y^3}{x^2+3xy+2y^2}$

$B=\dfrac{x^2\left(x+2y\right)-y^2\left(x+2y\right)}{x^2+xy+2xy+2y^2}$

$B=\dfrac{\left(x+2y\right)\left(x^2-y^2\right)}{x\left(x+y\right)+2y\left(x+y\right)}$

$B=\dfrac{\left(x+2y\right)\left(x-y\right)\left(x+y\right)}{\left(x+y\right)\left(2y+x\right)}$

$B=x-y$$\left(\text{Đ}K:x+2y\ne0;x+y\ne0\right)$

Tham khảo nhé~Câu trả lời: $B=\dfrac{x^3+2x^2y-xy^2-2y^3}{x^2+3xy+2y^2}$

$B=\dfrac{x^2\left(x+2y\right)-y^2\left(x+2y\right)}{x^2+xy+2xy+2y^2}$

$B=\dfrac{\left(x+2y\right)\left(x^2-y^2\right)}{x\left(x+y\right)+2y\left(x+y\right)}$

$B=\dfrac{\left(x+2y\right)\left(x-y\right)\left(x+y\right)}{\left(x+y\right)\left(2y+x\right)}$

$B=x-y$$\left(\text{Đ}K:x+2y\ne0;x+y\ne0\right)$

Tham khảo nhé~

Khôi Bùi · Answer

$B=\dfrac{x^3+2x^2y-xy^2-2y^3}{x^2+3xy+2y^2}$

$=\dfrac{x^2\left(x+2y\right)-y^2\left(x+2y\right)}{x^2+xy+2xy+2y^2}$

$=\dfrac{\left(x^2-y^2\right)\left(x+2y\right)}{x\left(x+y\right)+2y\left(x+y\right)}$

$=\dfrac{\left(x-y\right)\left(x+y\right)\left(x+2y\right)}{\left(x+2y\right)\left(x+y\right)}$

$=x-y$Câu trả lời: $B=\dfrac{x^3+2x^2y-xy^2-2y^3}{x^2+3xy+2y^2}$

$=\dfrac{x^2\left(x+2y\right)-y^2\left(x+2y\right)}{x^2+xy+2xy+2y^2}$

$=\dfrac{\left(x^2-y^2\right)\left(x+2y\right)}{x\left(x+y\right)+2y\left(x+y\right)}$

$=\dfrac{\left(x-y\right)\left(x+y\right)\left(x+2y\right)}{\left(x+2y\right)\left(x+y\right)}$

$=x-y$