`A` `=` $\dfrac{12+\dfrac{12}{13}-\dfrac{12}{17}}{4-\dfrac{4}{17}+\dfrac{4}{13}}$ `:` $\dfrac{\dfrac{3}{19}+\dfrac{3}{21}}{\dfrac{91}{19}+\dfrac{91}{21}}$ `.` $\dfrac{40.50+(-1)^{2}}{40+50+(-1)^{2}}$
`=` $\dfrac{12.(1+\dfrac{1}{13}-\dfrac{1}{17})}{4.(1-\dfrac{1}{17}+\dfrac{1}{13})}$ `:` $\dfrac{3.(\dfrac{1}{19}+\dfrac{1}{21})}{91.(\dfrac{1}{19}+\dfrac{1}{21})}$ `.` $\dfrac{2000+1}{90+1}$
`=` $\dfrac{12}{4}$ `:` $\dfrac{3}{91}$ `.` $\dfrac{2001}{91}$
`=` `3.` $\dfrac{91}{3}$ `.` $\dfrac{2001}{91}$
`=` $\dfrac{3.91.2001}{3.91}$
`=` `2001`
Vậy `A=2001`
