Question

Chứng minh đẳng thức :\(\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{\left(x+y\right)^2}}=\left|\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{1}{x+y}\right|\)

với \(x\ne o,y\ne0,x+y\ne0\)

GIẢI CHI TIẾT DÙM NHA CHI TIẾT ĐÓ!!!

Answer 1

\(\dfrac{2}{xy}-\dfrac{2}{y\left(x+y\right)}-\dfrac{2}{x\left(x+y\right)}=\dfrac{2\left(x+y\right)-2x-2y}{xy\left(x+y\right)}=0\)

\(A=\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{\left(x+y\right)^2}}\)

\(=\sqrt{\left(\dfrac{1}{x}\right)^2+\left(\dfrac{1}{y}\right)^2+\left(\dfrac{1}{x+y}\right)^2+2\times\dfrac{1}{x}\times\dfrac{1}{y}-2\times\dfrac{1}{y}\times\dfrac{1}{x+y}-2\times\dfrac{1}{x}\times\dfrac{1}{x+y}}\)

\(=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{1}{x+y}\right)}\)

\(=\left|\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{1}{x+y}\right|\left(\text{đ}pcm\right)\)