Câu hỏi của Lê Vương Kim Anh

Question

Thực hiện phép tính sau:

$\dfrac{x^2}{\left(x-y\right)^2\left(x+y\right)}-\dfrac{2x^2y^2}{x^4-2x^2y^2+y^4}+\dfrac{y^2}{\left(x^2-y^2\right)\left(x+y\right)}$

Phương Trâm · Accepted Answer

MTC: $\left(x-y\right)^2\left(x+y\right)^2$

$\dfrac{x^2}{\left(x-y\right)^2\left(x+y\right)}-\dfrac{2x^2y^2}{x^4-2x^2y^2+y^4}+\dfrac{y^2}{\left(x^2-y^2\right)\left(x+y\right)}$

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

$=\dfrac{x^3+x^2y-2xy^2+y^2x-y^3}{\left(x-y\right)^2\left(x+y\right)^2}$

$=\dfrac{x^3+x^2y-xy^2-y^3}{\left(x-y\right)^2\left(x+y\right)^2}$

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

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

$=\dfrac{1}{x-y}$Câu trả lời: MTC: $\left(x-y\right)^2\left(x+y\right)^2$

$\dfrac{x^2}{\left(x-y\right)^2\left(x+y\right)}-\dfrac{2x^2y^2}{x^4-2x^2y^2+y^4}+\dfrac{y^2}{\left(x^2-y^2\right)\left(x+y\right)}$

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

$=\dfrac{x^3+x^2y-2xy^2+y^2x-y^3}{\left(x-y\right)^2\left(x+y\right)^2}$

$=\dfrac{x^3+x^2y-xy^2-y^3}{\left(x-y\right)^2\left(x+y\right)^2}$

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

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

$=\dfrac{1}{x-y}$

Bài 6: Phép trừ các phân thức đại số