Question

[Đẹp Trai Không Bao Giờ Sai]

Cho a, b, c là các số thực khác 0. Tìm các số thực x, y, z khác 0 thỏa mãn

\(\dfrac{xy}{ay+bx}=\dfrac{yz}{bz+cy}=\dfrac{zx}{cx+az}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\)

Answer 1

[Nguyễn Thanh Hiền]

\(\dfrac{xy}{ay+bx}=\dfrac{yz}{bz+cy}=\dfrac{zx}{cx+az}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\left(1\right)\)

Ta có: \(\dfrac{xy}{ay+bx}=\dfrac{yz}{bz+cy}=\dfrac{zx}{cx+az}\)

\(\Rightarrow\dfrac{xyz}{ayz+bxz}=\dfrac{xyz}{bxz+cxy}=\dfrac{xyz}{cxy+ayz}\)

\(\Rightarrow ayz+bxz=bxz+cxy=cxy+ayz\)

\(\Rightarrow\left\{{}\begin{matrix}ayz+bxz=bxz+cxy\\ayz+bxz=cxy+ayz\\bxz+cxy=cxy+ayz\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}ayz=cxy\\bxz=cxy\\bxz=ayz\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}az=cx\\bz=cy\\bx=ay\end{matrix}\right.\left(2\right)\)

Thay (2) vào (1) ta có :

\(\dfrac{xy}{2ay}=\dfrac{yz}{2bz}=\dfrac{xz}{2cx}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\)

\(\Rightarrow\dfrac{x}{2a}=\dfrac{y}{2b}=\dfrac{z}{2c}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\left(3\right)\)

\(\Rightarrow\dfrac{x^2}{4a^2}=\dfrac{y^2}{4b^2}=\dfrac{z^2}{4c^2}=\dfrac{\left(x^2+y^2+z^2\right)^2}{\left(a^2+b^2+c^2\right)^2}=\)\(\dfrac{x^2+y^2+z^2}{4a^2+4b^2+4c^2}\)

\(\Rightarrow\dfrac{x^2+y^2+y^2}{a^2+b^2+c^2}=\dfrac{1}{4}\left(4\right).\)Thay (3) vào (2) ta có :

\(\dfrac{x}{2a}=\dfrac{y}{2b}=\dfrac{z}{2c}=\dfrac{1}{4}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{a}{2}\\y=\dfrac{b}{2}\\z=\dfrac{c}{2}\end{matrix}\right.\)

Answer 2

[Hoàng Quỳnh Phương]

CD+CH+CA=3C