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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
Ta có : \(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)
\(\Leftrightarrow\frac{abz-cya}{a^2}=\frac{bcx-baz}{b^2}=\frac{cay-cbx}{c^2}=\frac{abz-cyz+bcx-baz+cay-cbx}{a^2+b^2+c^2}\)
\(=\frac{0}{a^2+b^2+c^2}=0\)
\(\Rightarrow\hept{\begin{cases}bz=cy\\cx=az\\ay=bx\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{x}{c}=\frac{y}{b}\\\frac{x}{a}=\frac{z}{c}\\\frac{y}{b}=\frac{x}{a}\end{cases}}\Leftrightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)
\(\Rightarrow\frac{a\left(bz-cy\right)}{a^2}=\frac{b\left(cx-az\right)}{b^2}=\frac{c\left(ay-bx\right)}{c^2}\)
\(\Rightarrow\frac{abz-acy}{a^2}=\frac{bcx-baz}{b^2}=\frac{cay-cbx}{c^2}\)
Do a,b,c khác 0, áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\Rightarrow\frac{abz-acy}{a^2}=\frac{bcx-baz}{b^2}=\frac{cay-cbx}{c^2}=\frac{0}{a^2+b^2+c^2}=0\)
\(\hept{\begin{cases}bz-cy=0\\cx-az=0\\ay-bx=0\end{cases}\Rightarrow\hept{\begin{cases}\frac{y}{b}=\frac{z}{c}\\\frac{x}{a}=\frac{z}{c}\\\frac{x}{a}=\frac{y}{b}\end{cases}\Rightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}}}\)
Ta có: \(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}\)
\(\Leftrightarrow\frac{x\left(bz-cy\right)}{ax}=\frac{y\left(cx-az\right)}{by}=\frac{z\left(ay-bx\right)}{cz}\)
\(\Leftrightarrow\frac{bxz-cxy}{ax}=\frac{cxy-ayz}{by}=\frac{ayz-bxz}{cz}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\Leftrightarrow\frac{bxz-cxy+cxy-ayz+ayz-bxz}{ax+by+cz}=\frac{0}{ax+by+cz}=0\)
\(\Rightarrow\begin{cases}\frac{bz-cy}{a}=0\Leftrightarrow bz-cy=0\Leftrightarrow bz=cy\Leftrightarrow\frac{z}{c}=\frac{y}{b}\left(1\right)\\\frac{cx-az}{b}=0\Leftrightarrow cx-az=0\Leftrightarrow cx=az\Leftrightarrow\frac{x}{a}=\frac{z}{c}\left(2\right)\\\frac{ay-bx}{c}=0\Leftrightarrow ay-bx=0\Leftrightarrow ay=bx\Leftrightarrow\frac{y}{b}=\frac{x}{a}\left(3\right)\end{cases}\)
Từ (1),(2),(3) suy ra \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\)