21/03/2025
21/03/2025
21/03/2025
Đặt
$\displaystyle \frac{bp-cn}{a} =\frac{cm-ap}{b} =\frac{an-bm}{c} =k\ \ \ \ ( a,b,c\neq 0)$
⟹ $\displaystyle \begin{array}{{>{\displaystyle}l}}
\frac{1}{bc} .\frac{bp-cn}{a} =\frac{1}{bc} .\frac{ak}{a} \Longrightarrow \frac{1}{a} .\frac{bp-cn}{bc} =\frac{1}{a} .\frac{ak}{bc} \Longrightarrow \frac{p}{c} -\frac{n}{b} =\frac{ak}{bc}\\
\frac{1}{ac} .\frac{cm-ap}{b} =\frac{1}{ac} .\frac{bk}{b} \Longrightarrow \frac{1}{b} .\frac{cm-ap}{ac} =\frac{1}{b} .\frac{bk}{ac} \Longrightarrow \frac{m}{a} -\frac{p}{c} =\frac{bk}{ac}\\
\frac{1}{ab} .\frac{an-bm}{c} =\frac{1}{ab} .\frac{ck}{c} \Longrightarrow \frac{1}{c} .\frac{an-bm}{ab} =\frac{1}{c} .\frac{ck}{ab} \Longrightarrow \frac{n}{b} -\frac{m}{a} =\frac{ck}{ab}
\end{array}$
⟹ $ $\displaystyle \frac{p}{c} -\frac{n}{b} +\frac{m}{a} -\frac{p}{c} +\frac{n}{b} -\frac{m}{a} =\frac{ak}{bc} +\frac{bk}{ac} +\frac{ck}{ab}$
⟹ $\displaystyle \left(\frac{p}{c} -\frac{p}{c}\right) +\left(\frac{n}{b} -\frac{n}{b}\right) +\left(\frac{m}{a} -\frac{m}{a}\right) =k\left(\frac{a}{bc} +\frac{b}{ac} +\frac{c}{ab}\right)$
⟹ $\displaystyle k\left(\frac{a}{bc} +\frac{b}{ac} +\frac{c}{ab}\right) =0$
⟹ $\displaystyle k=0$ $\displaystyle \ \ Vì\ \left(\frac{a}{bc} +\frac{b}{ac} +\frac{c}{ab} =\frac{a^{2} +b^{2} +c^{2}}{abc} \neq 0\ \ \forall a,b,c\neq 0\right)$
⟹ $\displaystyle bp-cn=cm-ap=an-bm=0$
⟹ $\displaystyle \begin{array}{{>{\displaystyle}l}}
bp=cn\Longrightarrow \frac{n}{b} =\frac{p}{c}\\
cm=ap\Longrightarrow \frac{p}{c} =\frac{m}{a}
\end{array}$
⟹$\displaystyle \frac{m}{a} =\frac{n}{b} =\frac{p}{c}$ (dpcm)
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