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Giải thích các bước giải:

Ta có:

$\dfrac{3a+2b+c}{a+2b-c}=\dfrac{3a-2b+c}{a-2b-c}$

$\to \dfrac{(a+2b-c)+(2a+2c)}{a+2b-c}=\dfrac{(a-2b-c)+(2a+2c)}{a-2b-c}$

$\to 1+\dfrac{2a+2c}{a+2b-c}=1+\dfrac{2a+2c}{a-2b-c}$

$\to \dfrac{2a+2c}{a+2b-c}=\dfrac{2a+2c}{a-2b-c}$

$\to \dfrac{2(a+c)}{a+2b-c}=\dfrac{2(a+c)}{a-2b-c}(1)$

Vì $b\ne 0\to 4b\ne 0$

$\to 2b\ne -2b$

$\to a-c+2b\ne a-c-2b$

$\to a+2b-c\ne a-2b-c$

Kết hợp $(1)$

$\to 2(a+c)=0$

$\to a+c=0$

$\to đpcm$