Question

[❤Hàn Tử Thiên❤]

Chứng minh rằng nếu : \(\frac{a+b}{b+c}=\frac{c+d}{d+a}\) thì a = c hoặc a + b + c + d = 0

Answer 1

[❤♚ℳℴℴทℛℴƴຮ♚❤]

Ta có:\(\frac{a+b}{b+c}=\frac{c+d}{d+a}\)

\(\implies\)\(\frac{a+b}{c+d}=\frac{b+c}{d+a}\)

\(\implies\) \(\frac{a+b}{c+d}+1=\frac{b+c}{d+a}+1\)

\(\implies\) \(\frac{a+b+c+d}{c+d}=\frac{a+b+c+d}{d+a}\)

\(\implies\) \(\frac{a+b+c+d}{c+d}-\frac{a+b+c+d}{d+a}=0\)

\(\implies\) \(\left(a+b+c+d\right)\left(\frac{1}{c+d}-\frac{1}{d+a}\right)=0\)

\(\implies\)\(\orbr{\begin{cases}a+b+c+d=0\\\frac{1}{c+d}-\frac{1}{d+a}=0\end{cases}}\)

\(\implies\) \(\orbr{\begin{cases}a+b+c+d=0\\\frac{1}{c+d}=\frac{1}{d+a}\end{cases}}\)

\(\implies\) \(\orbr{\begin{cases}a+b+c+d=0\\c+d=d+a\end{cases}}\)

\(\implies\) \(\orbr{\begin{cases}a+b+c+d=0\\c=a\end{cases}}\)

Answer 2

[IS]

ta có \(\frac{a+b}{b+c}=\frac{c+d}{d+a}\)

=>\(\left(a+b\right)\left(a+d\right)=\left(c+d\right)\left(b+c\right)\)

=> \(a^2+ab+ad+bd=c^2+bc+bd+cd\)

=>\(a^2+ab+ad-bc-c^2-cd=0\)

=>\(\left(a^2-c^2\right)+\left(ab-cd\right)+\left(ab-ac\right)=0\)

=>\(\left(a-c\right)\left(a+c\right)+d\left(a-c\right)+b\left(a-c\right)=0\)

=>\(\left(a-c\right)\left(a+b+c+d\right)=0\)

=>\(\orbr{\begin{cases}a-c=0\\a+b+c+d=0\end{cases}\left(dpcm\right)}\)

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