Układ dwóch równań:
3=a+b
4=√2*a+b
Proszę o podanie ile wynosi "a" i "b"
p(2) - pierwiastek kwadratowy z 2
3 = a + b
4 = p(2) a + b
---------------------- odejmujemy stronami
4 - 3 = p(2)a - a
1 = ( p(2) - 1) a
a = 1 /[ p(2) - 1] = [ 1*( p(2) + 1)]/[ ( p(2) - 1)*( p(2) + 1)] = p(2) + 1
a = p(2) + 1
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b = 3 - a = 3 - ( p(2) + 1) = 2 - p(2)
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p(2) - pierwiastek kwadratowy z 2
3 = a + b
4 = p(2) a + b
---------------------- odejmujemy stronami
4 - 3 = p(2)a - a
1 = ( p(2) - 1) a
a = 1 /[ p(2) - 1] = [ 1*( p(2) + 1)]/[ ( p(2) - 1)*( p(2) + 1)] = p(2) + 1
a = p(2) + 1
==========
b = 3 - a = 3 - ( p(2) + 1) = 2 - p(2)
==================================