Question

Tìm các số x,y,z khác 0 biết: \(\frac{xy}{ay+bx}=\frac{yz}{cy+bz}=\frac{xz}{az+cx}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}\left(a,b,c\ne0\right)\)

Answer 1

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

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

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Answer 2

ay+bxxy​=bz+cyyz​=cx+azxz​=a2+b2+c2x2+y2+z2​(1)

Ta có: xyay+bx=yzbz+cy=xzcx+az.ay+bxxy​=bz+cyyz​=cx+azxz​.

⇒xyzayz+bxz=xyzbxz+cxy=xyzcxy+ayz.⇒ayz+bxzxyz​=bxz+cxyxyz​=cxy+ayzxyz​.

$\Rightarrow ayz+bxz=bxz+cxy=cxy+ayz$

⇒{ayz+bxz=bxz+cxyayz+bxz=cxy+ayzbxz+cxy=cxy+ayz⇒{ayz=cxybxz=cxybxz=ayz⇒{az=cxbz=cybx=ay(2)⇒⎩⎨⎧​ayz+bxz=bxz+cxyayz+bxz=cxy+ayzbxz+cxy=cxy+ayz​⇒⎩⎨⎧​ayz=cxybxz=cxybxz=ayz​⇒⎩⎨⎧​az=cxbz=cybx=ay​(2)

Thay (2) vào (1) ta được:

xyay+ay=yzbz+bz=xzcx+cxay+ayxy​=bz+bzyz​=cx+cxxz​

⇒xy2ay=yz2bz=xz2cx=x2+y2+z2a2+b2+c2⇒2ayxy​=2bzyz​=2cxxz​=a2+b2+c2x2+y2+z2​

⇒x2a=y2b=z2c=x2+y2+z2a2+b2+c2(3).⇒2ax​=2by​=2cz​=a2+b2+c2x2+y2+z2​(3).

⇒x24a2=y24b2=z24c2=(x2+y2+z2)2(a2+b2+c2)2=x2+y2+z24a2+4b2+4c2⇒4a2x2​=4b2y2​=4c2z2​=(a2+b2+c2)2(x2+y2+z2)2​=4a2+4b2+4c2x2+y2+z2​

⇒x2+y2+z24a2+4b2+4c2=1.(x2+y2+z2)4.(a2+b2+c2)⇒4a2+4b2+4c2x2+y2+z2​=4.(a2+b2+c2)1.(x2+y2+z2)​

⇒x2+y2+z2a2+b2+c2=14(4).⇒a2+b2+c2x2+y2+z2​=41​(4).

Từ (3) và (4)

⇒x2a=y2b=z2c=14.⇒2ax​=2by​=2cz​=41​.

⇒{x2a=14y2b=14z2c=14⇒{x=14.2ay=14.2bz=14.2c⇒{x=a2y=b2z=c2⇒⎩⎨⎧​2ax​=41​2by​=41​2cz​=41​​⇒⎩⎨⎧​x=41​.2ay=41​.2bz=41​.2c​⇒⎩⎨⎧​x=2a​y=2b​z=2c​​

Vậy x=a2;y=b2;z=c2(x,y,z≠0);(a,b,c≠0).x=2a​;y=2b​;z=2c​(x,y,z=0);(a,b,c=0).

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