câu hỏi tương tự nhé 

Vào CHTT, mình chỉ bít mỗi cách ấy thôi.

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

Ta có: \(\frac{xy}{ay+bx}=\frac{yz}{bz+cy}=\frac{xz}{cx+az}.\)

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

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

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

Thay (2) vào (1) ta được:

\(\frac{xy}{ay+ay}=\frac{yz}{bz+bz}=\frac{xz}{cx+cx}\)

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

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

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

\(\Rightarrow\frac{x^2+y^2+z^2}{4a^2+4b^2+4c^2}=\frac{1.\left(x^2+y^2+z^2\right)}{4.\left(a^2+b^2+c^2\right)}\)

\(\Rightarrow\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=\frac{1}{4}\left(4\right).\)

Từ (3) và (4)

\(\Rightarrow\frac{x}{2a}=\frac{y}{2b}=\frac{z}{2c}=\frac{1}{4}.\)

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

Vậy \(x=\frac{a}{2};y=\frac{b}{2};z=\frac{c}{2}\left(x,y,z\ne0\right);\left(a,b,c\ne0\right).\)

Chúc bạn học tốt!

\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\) => \(\frac{a.\left(bz-cy\right)}{a^2}=\frac{b.\left(cx-az\right)}{b^2}=\frac{c.\left(ay-bx\right)}{c^2}\)

<=> \(\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{cay-bcx}{c^2}\). Theo tính chất dãy tỉ số bằng nhau

=> \(\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{cay-bcx}{c^2}=\frac{abz-acy+bcx-abz+cay-bcx}{a^2+b^2+c^2}=0\)

=> \(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\) = 0 

=> \(bz-cy=0\Rightarrow bz=cy\Rightarrow\frac{y}{b}=\frac{z}{c}\) (1)

\(cx-az=0\Rightarrow\frac{x}{a}=\frac{z}{c}\)  (2)

Từ (1)(2) => \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)

Ta có : \(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)

\(\Leftrightarrow\frac{abz-cya}{a^2}=\frac{bcx-baz}{b^2}=\frac{cay-cbx}{c^2}=\frac{abz-cyz+bcx-baz+cay-cbx}{a^2+b^2+c^2}\)

\(=\frac{0}{a^2+b^2+c^2}=0\)

\(\Rightarrow\hept{\begin{cases}bz=cy\\cx=az\\ay=bx\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{x}{c}=\frac{y}{b}\\\frac{x}{a}=\frac{z}{c}\\\frac{y}{b}=\frac{x}{a}\end{cases}}\Leftrightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)