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

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 :

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

CD+CH+CA=3C

Phương Ann Nhã Doanh đề bài khó wá Mashiro Shiina Đinh Đức Hùng

Nguyễn Huy Tú Lightning Farron Akai Haruma

Lời giải:
Sửa đề: $z$ đầu tiên ở mẫu đổi thành $a$.

Ta có:

$\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}$

$=\frac{abz-cya}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}$

$=\frac{abz-cya+bcx-abz+acy-bcx}{a^2+b^2+c^2}=\frac{0}{a^2+b^2+c^2}=0$

$\Rightarrow bz-cy=cx-az=ay-bx=0$

$\Rightarrow bz=cy; cx=az; ay=bx$

$\Rightarrow \frac{x}{a}=\frac{y}{b}=\frac{z}{c}$

Ta có đpcm.

Ta có :

\(\dfrac{cy-bx}{x}=\dfrac{az-cx}{y}=\dfrac{bx-ay}{z}=\dfrac{bxz-cxy+cxy-ayz+ayz-bxz}{ax+by+cz}=0\)

\(\Rightarrow\dfrac{cy-bz}{x}=0\) \(\Rightarrow cy=bz\) \(\Rightarrow\) \(\dfrac{b}{y}=\dfrac{c}{z}\left(1\right)\)

\(\Rightarrow\dfrac{az-cx}{y}=0\) \(\Rightarrow az=cx\) \(\Rightarrow\dfrac{a}{x}=\dfrac{c}{z}\left(2\right)\)

Từ (1) và (2) suy ra : \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)

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

\(\Rightarrow\dfrac{a\left(bz-cy\right)}{a^2}\text{=}\dfrac{b\left(cx-az\right)}{b^2}\text{=}\dfrac{c\left(ay-bx\right)}{c^2}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có :

\(\dfrac{a\left(bz-cy\right)}{a^2}\text{=}\dfrac{b\left(cx-az\right)}{b^2}\text{=}\dfrac{c\left(ay-bx\right)}{c^2}\text{=}\dfrac{abz-acy+bcz-baz+cay-cbx}{a^2+b^2+c^2}\text{=}0\)

\(\Rightarrow\dfrac{bz-cy}{a}\text{=}0\Rightarrow bz\text{=}cy\)

\(\Rightarrow\dfrac{b}{c}\text{=}\dfrac{y}{z}\left(1\right)\)

\(\dfrac{cx-az}{b}\text{=}0\Rightarrow cx\text{=}az\)

\(\Rightarrow\dfrac{c}{a}\text{=}\dfrac{z}{x}\left(2\right)\)

Từ (1) và (2):

\(\Rightarrow dpcm\)