Ta có : \(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\Leftrightarrow\frac{baz-cay}{a^2}=\frac{cbx-abz}{b^2}=\frac{acy-bcx}{c^2}=\frac{baz-cay+cbx-abz+acy-bcx}{a^2+b^2+c^2}=0\)

\(\Rightarrow bz=cy\Leftrightarrow\frac{y}{b}=\frac{z}{c}\)

\(\Rightarrow cx=az\Leftrightarrow\frac{x}{a}=\frac{z}{c}\) 

\(\Rightarrow ay=bx\Leftrightarrow\frac{x}{a}=\frac{y}{b}\)

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

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

\(\Leftrightarrow\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}\)

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

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

\(\Leftrightarrow x:y:z=a:b:c\)

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

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

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

Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

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

=> \(\hept{\begin{cases}\frac{bz-cy}{a}=0\\\frac{cx-az}{b}=0\\\frac{ay-bx}{c}=0\end{cases}}\) => \(\hept{\begin{cases}bz-cy=0\\cx-az=0\\ay-bx=0\end{cases}}\) => \(\hept{\begin{cases}bz=cy\\cx=az\\ay=bx\end{cases}}\) => \(\hept{\begin{cases}\frac{b}{y}=\frac{c}{z}\\\frac{c}{z}=\frac{a}{x}\\\frac{a}{x}=\frac{b}{y}\end{cases}}\) => \(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)=> \(a:b:c=x:y:z\)

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

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

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

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

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

\(\Rightarrow\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}\)

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

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

\(\Rightarrow\hept{\begin{cases}bz-cy=0\\cx-az=0\\ay-bx=0\end{cases}}\Rightarrow\hept{\begin{cases}bz=cy\\cx=az\\ay=bx\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\frac{y}{b}=\frac{z}{c}\\\frac{x}{a}=\frac{z}{c}\\\frac{y}{b}=\frac{x}{a}\end{cases}}\Rightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)

* C1 :(bz - cy)/a = (abz - acy)/a²

(cx - az)/b = (bcx - abz)/b²

(ay - bx)/c = (acy - bcx)/c²

Mà (bz - cy)/a = (cx - az)/b = (ay - bx)/c

=>(abz - acy)/a² = (bcx - abz)/b² = (acy - bcx)/c² = (abz - acy + bcx - abz + acy - bcx)/a² + b² + c² = 0

=>(bz - cy)/a = (cx - az)/b = (ay - bx)/c = 0

=>bz - cy = cx - az = ay - bx = 0

*Xét bz - cy = 0

=>bz = cy

=>z/c = y/b

Chứng minh tương tự = >x/a = y/b ; x/a = z/c

=> x/a = y/b = z/c

*C2 : 

(bz - cy)/a = (abz - acy)/ax

(cx - az)/by = (bcx - abz)/by

(ay - bx)/cz = (acy - bcx)/cz

Làm tương tự như C1

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

Suy ra: \(\frac{bxz-cxy}{ax}=\frac{cxy-azy}{bx}=\frac{azy-bxz}{cx}\). Áp dụng tính chất dãy tỉ số bằng nhau có: 

\(\frac{bxz-cxy}{ax}=\frac{cxy-azy}{bx}=\frac{azy-bxz}{cx}=\frac{\left(bxz-cxy\right)+\left(cxy-azy\right)+\left(azy-bxz\right)}{ax+bx+cx}\)

\(=\frac{\left(bxz-bxz\right)-\left(cxy-cxy\right)-\left(azy-azy\right)}{ax+by+cz}=\frac{0}{ax+by+cz}\)

Suy ra: \(\hept{\begin{cases}bz-cy=0\\cx-az=0\\ay-bx=0\end{cases}\Leftrightarrow}\hept{\begin{cases}bz=cy\\cx=az\\ay=bx\end{cases}}\)

Áp dụng tính chất tỉ lệ thức ta được: \(\hept{\begin{cases}\frac{y}{b}=\frac{z}{c}\\\frac{z}{c}=\frac{x}{a}\\\frac{x}{a}=\frac{y}{b}\end{cases}\Leftrightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}^{\left(đpcm\right)}}\)

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

\(\Rightarrow\frac{abz-acy}{a^2}=\frac{bcx-baz}{b^2}=\frac{cay-cbx}{c^2}\)

Do a,b,c khác 0, áp dụng tính chất dãy tỉ số bằng nhau ta có: 

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

\(\hept{\begin{cases}bz-cy=0\\cx-az=0\\ay-bx=0\end{cases}\Rightarrow\hept{\begin{cases}\frac{y}{b}=\frac{z}{c}\\\frac{x}{a}=\frac{z}{c}\\\frac{x}{a}=\frac{y}{b}\end{cases}\Rightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}}}\)

áp dụng tính chất hai dãy tỉ số bằng nhau nha bạn