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

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

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

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

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

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

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

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

suy ra:

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

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

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

Chứng minh x,y,z = a,b,y là sao ? Là x : y : z = a : b : y hay thế nào ?

x,y,z = a,b,y là gì vậy?

Có: Đề \(\Leftrightarrow\frac{abz-acy}{a^2}=\frac{bcx-abz}{b^2}=\frac{acy-bcx}{c^2}\)\(=\frac{\left(abz-abz\right)+\left(bcx-bcx\right)+\left(acy-acy\right)}{a^2+b^2+c^2}\)

\(=\frac{0}{a^2+b^2+c^2}=0\)\(\left(ĐKXĐ:a,b,c\ne0\right)\)

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

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

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

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

=> bz - cy = cx - az = ay - bx = 0

+) bz - cy = 0 => bz = cy => y/b = z/c

+) cx - az = 0 => cx = az => x/a = z/c

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

vế 1 thiếu x

vế 2 thiếu y

vế 3 thiếu z

nhấn ba vế với cái thiếu

ta có

\(\frac{bxz-cxy}{ax}=\frac{cxy-ayz}{by}=\frac{ayz-bxy}{cz}\)

Theo TCDTSBN`, ta có

\(\frac{bxz-cxy}{ax}=\frac{cxy-ayz}{by}=\frac{ayz-bxy}{cz}\)

= cộng chừng đó lại tử + tử, mẫu + mẫu

=0/(ax+by+cz)

=0

=>bzx=cxy

=>cxy=ayz

=>bxz=cxy=ayz

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

đó mỏi tay lắm rồi đó