Question

rút gọn bt {[1/(x^2+2xy+y^2)]-[1/(x^2-y^2)]}/(4xy/y^2-x^2)

Answer 1

\(\dfrac{1}{x^2+2xy+y^2}-\dfrac{1}{x^2-y^2}:\dfrac{4xy}{y^2-x^2}\) \(\left(x,y\ne0;x\ne\pm y\right)\)

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

\(=\dfrac{1}{x^2+2xy+y^2}+\dfrac{1}{4xy}\)

\(=\dfrac{6xy+x^2+y^2}{4xy\left(x+y\right)^2}\)

Answer 2

Ta có: \(\dfrac{1}{x^2+2xy+y^2}-\dfrac{1}{x^2-y^2}:\dfrac{4xy}{y^2-x^2}\)

\(=\dfrac{1}{\left(x+y\right)^2}+\dfrac{1}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\left(x+y\right)\left(x-y\right)}{4xy}\)

\(=\dfrac{1}{\left(x+y\right)^2}+\dfrac{1}{4xy}\)

\(=\dfrac{4xy}{4xy\left(x+y\right)^2}+\dfrac{x^2+2xy+y^2}{4xy\left(x+y\right)^2}\)

\(=\dfrac{x^2+6xy+y^2}{4xy\left(x+y\right)^2}\)