Cho A = 1/(x+y)^3 (1/x^4 -1/y^4), B = 2/(x+y)^4 (1/x^3 - 1/y^3)

* Hướng dẫn giải

$A=\frac{y^4-x^4}{x^4{y}^4{\left(x+y\right)}^3}=\frac{\left(y^2+x^2\right)\left(y^2-x^2\right)}{x^4{y}^4{\left(x+y\right)}^3}=\frac{\left(y^2+x^2\right)\left(y-x\right)}{x^4{y}^4{\left(x+y\right)}^2}.$

$B+C=\frac{2}{{\left(x+y\right)}^4}\left(\frac{1}{x^3}-\frac{1}{y^3}+\frac{1}{x+y}.\frac{y^2-x^2}{x^2{y}^2}\right)$

$=\frac{2}{{\left(x+y\right)}^4}\left(\frac{1}{x^3}-\frac{1}{y^3}+\frac{y-x}{x^2{y}^2}\right)=\frac{2}{{\left(x+y\right)}^4}.\frac{y^3-x^3+xy\left(y-x\right)}{x^3{y}^3}$

$=\frac{2}{{\left(x+y\right)}^4}.\frac{\left(y-x\right)\left(y^2+2xy+x^2\right)}{x^3{y}^3}=\frac{2\left(y-x\right)}{{\left(x+y\right)}^2{x}^3{y}^3}.$

Do đó $A+B+C=\frac{\left(y^2+x^2\right)\left(y-x\right)}{x^4{y}^4{\left(x+y\right)}^2}+\frac{2\left(y-x\right)}{{\left(x+y\right)}^2{x}^3{y}^3}$

$=\frac{\left(y^2+x^2\right)\left(y-x\right)+2xy\left(y-x\right)}{x^4{y}^4{\left(x+y\right)}^2}=\frac{\left(y-x\right)\left(y^2+x^2+2xy\right)}{x^4{y}^4{\left(x+y\right)}^2}=\frac{y-x}{x^4{y}^4}$