Ta có:
`@`\(a+b-c=\dfrac{1}{2021}\) \(\Rightarrow\dfrac{1}{a+b-c}=2021\)
`@`\(\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{c}=2021\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{c}=\dfrac{1}{a+b-c}\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{c}-\dfrac{1}{a+b-c}=0\)
\(\Leftrightarrow\dfrac{a+b}{ab}-\dfrac{a+b-c+c}{c\left(a+b-c\right)}=0\)
\(\Leftrightarrow\dfrac{a+b}{ab}-\dfrac{a+b}{c\left(a+b-c\right)}=0\)
\(\Leftrightarrow\left(a+b\right)\left[\dfrac{1}{ab}-\dfrac{1}{c\left(a+b-c\right)}\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left[\dfrac{c\left(a+b-c\right)-ab}{abc\left(a+b-c\right)}\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left[\dfrac{ac+bc-c^2-ab}{abc\left(a+b-c\right)}\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left[\dfrac{c\left(a-c\right)-b\left(a-c\right)}{abc\left(a+b-c\right)}\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left[\dfrac{\left(a-c\right)\left(c-b\right)}{abc\left(a+b-c\right)}\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left(a-c\right)\left(c-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b=0\\a-c=0\\c-b=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a=-b\\a=c\\c=b\end{matrix}\right.\)
\(A=\left(a^{2021}+b^{2021}-c^{2021}\right).\left(\dfrac{1}{a^{2021}}+\dfrac{1}{b^{2021}}-\dfrac{1}{c^{2021}}\right)\)
\(A=\left(a^{2021}+a^{2021}-a^{2021}\right).\left(\dfrac{1}{a^{2021}}+\dfrac{1}{a^{2021}}-\dfrac{1}{a^{2021}}\right)\)
\(A=a^{2021}.\dfrac{1}{a^{2021}}\)
\(A=1\)
Vậy \(A=1\)
