Ile jesr par liczb naturalnych (x,y) takich że x2−y2 = 48 ma wyjść 3 . prosze o dokładne obliczenie !!
x^2 - y^2 = 48
( x - y)*( x + y) = 48
( x - y)*( x + y) = 2*24 = 6*8 = 4*12
zatem
x - y = 2 lub x - y = 6 lub x - y = 4
x + y = 24 x + y = 8 x + y = 12
więc
2 x = 26 lub 2 x = 14 lub 2 x = 16
x = 13 x = 7 x = 8
y = 11 y = 1 y = 4
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Odp. x = 13 , y = 11 lub x = 7 , y = 1 lub x = 8, y = 4
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x^2 - y^2 = 48
( x - y)*( x + y) = 48
( x - y)*( x + y) = 2*24 = 6*8 = 4*12
zatem
x - y = 2 lub x - y = 6 lub x - y = 4
x + y = 24 x + y = 8 x + y = 12
więc
2 x = 26 lub 2 x = 14 lub 2 x = 16
x = 13 x = 7 x = 8
y = 11 y = 1 y = 4
====================================================
Odp. x = 13 , y = 11 lub x = 7 , y = 1 lub x = 8, y = 4