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Đáp án: $1$ 

Giải thích các bước giải:

Ta có:

$a+b=c+\dfrac7{2025}$

$\to a+b-c=\dfrac7{2025}$

$\dfrac7a+\dfrac{14}{2b}=\dfrac{21}{3c}+2025$

$\to \dfrac7a+\dfrac7b=\dfrac7c+2025$

$\to \dfrac1a+\dfrac1b=\dfrac1c+\dfrac{2025}7$

$\to \dfrac1a+\dfrac1b-\dfrac1c=\dfrac{2025}7$

$\to (\dfrac1a+\dfrac1b-\dfrac1c)\cdot (a+b-c)=1$

$\to (bc+ac-ab)(a+b-c)=abc$

$\to b^2c-bc^2+ca^2-c^2a-ba^2-b^2a+3bca=abc$

$\to b^2c-bc^2+ca^2-c^2a-ba^2-b^2a+2abc=0$

$\to bc(b-c)-a^2(b-c)-a(b^2-2bc+c^2)=0$

$\to bc(b-c)-a^2(b-c)-a(b-c)^2=0$

$\to (b-c)(bc-a^2-a(b-c))=0$

$\to (b-c)(c-a)(a+b)=0$

$\to b=c$ hoặc $c=a$ hoặc $a=-b$

$\to A=(a^{2025}+b^{2025}-c^{2025})\cdot (\dfrac1{a^{2025}}+\dfrac1{b^{2025}}-\dfrac1{c^{2025}})$

$\to 1$