Question

Cho \(\dfrac{x}{m} + \dfrac{y}{n} + \dfrac{z}{p} \) . Rút gọn A= \(\dfrac{x^2+y^2+z^2}{(mx+ny+pz)^2} \)

Answer 1

Đặt \(\dfrac{x}{m} + \dfrac{y}{n} + \dfrac{z}{p} = k\)

<=> \(\dfrac{x}{m} =k <=> x = mk \)

<=> \(\dfrac{y}{n} = k <=> y =nk\)

<=> \(\dfrac{z}{p} = k <=> z = pk\)

Thay \(x = mk ; y=nk ; z=pk\) vào A , ta có :

\(\dfrac{(mk)^2+(nk)^2+(pk)^2}{(m^2k+n^2+p^2k)^2}\)

= \(\dfrac{m^2k^2+n^2k^2+p^2k^2}{(m^4k^2+n^4k^2+p^4k^2+2m^2n^2k^2+2n^2p^2k^2+2m^2p^2k^2)}\)

= \(\dfrac{k^2(m^2+n^2+p^2}{k^2(m^4+n^4+p^4+2m^2n^2+2n^2p+2m^2p^2)}\)

= \(\dfrac{k^2(m^2+n^2+p^2}{k^2(m^2+n^2+p^2)^2}\)

= \(\dfrac{1}{m^2+n^2+p^2} \)

Vậy A = \(\dfrac{1}{m^2+n^2+p^2}\)