Sửa lại đề bài \(\dfrac{a-2c}{3a+c}=\dfrac{b-2d}{3b+d}\)
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a-2c}{b-2d}\left(1\right)\)
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{3a+c}{3b+d}\left(2\right)\)
\(\left(1\right);\left(2\right)\Rightarrow\dfrac{a-2c}{b-2d}=\dfrac{3a+b}{3b+d}\)
\(\Rightarrow\dfrac{a-2c}{3a+c}=\dfrac{b-2d}{3b+d}\left(đpcm\right)\)
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