may tinh casio

H

hiensau99

Ta có: $(A^3-3AB^2)^2=2004^2 \to A^6+9A^2B^4-6A^4B^2=2004^2 $ (1)

$(B^3-3A^2B)^2=2005^2 \to B^6-6A^2B^4+9A^4B^2=2005^2$ (2)

C[FONT=&quot]ộ[/FONT]ng v[FONT=&quot]ế[/FONT] v[FONT=&quot]ớ[/FONT]i v[FONT=&quot]ế[/FONT] c[FONT=&quot]ủ[/FONT]a (1) và (2) có:
$A^6+9A^2B^4-6A^4B^2+B^6-6A^2B^4+9A^4B^2=2004^2+2005^2$

$\to A^6+3A^2B^4+3A^4B^2+B^6=2004^2+2005^2$

$\to (A^2+B^2)^3=2004^2+2005^2 \to A^2+B^2 \approx 200,299 $



 
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