1 | 1 | |||
Wiadomo że x+ | =4. Oblicz x3+ | |||
x | x3 |
3 | 1 | |||
(x+1/x)3=x3+3x+ | + | =64 | ||
x | x3 |
1 | ||
x3+12+ | =64 | |
x3 |
1 | ||
x3+ | =52 | |
x3 |
1 | 1 | |||
x2 + | = (x+ | )2 − 2 = 16 − 2 = 14 | ||
x2 | x |
1 | 1 | 1 | 1 | |||||
x3 + | = (x+ | )(x2 − 1 + | ) = 4(x2+ | −1) = 4(14−1) =52 | ||||
x3 | x | x2 | x2 |
5^2 | 52 |
2^{10} | 210 |
a_2 | a2 |
a_{25} | a25 |
p{2} | √2 |
p{81} | √81 |
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