Question

[Vũ Xuân Hiếu]

Các bạn jup mk nha

Answer 1

[Linh_Windy]

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

\(\Rightarrow\dfrac{abz-acy}{a^2}=\dfrac{cbx-abz}{b^2}=\dfrac{acy-cbx}{c^2}\)

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

\(\dfrac{abz-acy}{a^2}=\dfrac{cbx-abz}{b^2}=\dfrac{acy-cbx}{c^2}\)

\(=\dfrac{abz-acy+cbx-abz+acy-cbx}{a^2+b^2+c^2}\)

\(=\dfrac{\left(abz-abz\right)+\left(acy-acy\right)+\left(cbx-cbx\right)}{a^2+b^2+c^2}\)

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

Nên:

\(\left\{{}\begin{matrix}bz=cy\\cx=az\\ay=bx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{y}=\dfrac{c}{z}\\\dfrac{a}{x}=\dfrac{c}{z}\\\dfrac{a}{x}=\dfrac{b}{y}\end{matrix}\right.\)

Suy ra: \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)

Hay \(x:y:z=a:b:c\left(đpcm\right)\)

Answer 2

[ChaosKiz]

Ta có:

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

\(\Leftrightarrow\dfrac{abz-acy}{a^2}=\dfrac{cbx-abz}{b^2}=\dfrac{acy-cbx}{c^2}\)

Áp dụng tính chất ta có:

\(\dfrac{abz-acy}{a^2}=\dfrac{cbx-abz}{b^2}=\dfrac{acy-cbx}{c^2}\)

\(=\dfrac{\left(abz-acy\right)+\left(cbx-abz\right)+\left(acy-cbx\right)}{a^2+b^2+c^2}\)

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

\(\Rightarrow\left\{{}\begin{matrix}bz=cy\\cx=az\\ay=bx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{y}=\dfrac{c}{z}\\\dfrac{a}{x}=\dfrac{c}{z}\\\dfrac{a}{x}=\dfrac{b}{y}\end{matrix}\right.\)

\(\Rightarrow\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)

hay \(x:y:z=a:b:c\left(đpcm\right)\)