1 | 1 | |||
Udowodnij, że jeżeli x+ | =3 to x3+ | =18 | ||
x | x3 |
1 | 1 | 1 | ||||
x3+ | =(x+ | )3−3(x+ | )= 27−3*3=... | |||
x3 | x | x |
1 | 1 | 1 | 1 | |||||
(x + | )3 = x3 + 3 • x2 • | + 3 • x • | + | = | ||||
x | x | x2 | x3 |
3 | 1 | 1 | 1 | |||||
x3 + 3x + | + | = x3 + | + 3(x + | ), więc skoro | ||||
x | x3 | x3 | x |
1 | 1 | 1 | ||||
(x + | )3 = x3 + | + 3(x + | ) to | |||
x | x3 | x |
1 | 1 | 1 | 1 | |||||
x3 + | = (x + | )3 − 3(x + | ), wiesz ile wynosi x + | , | ||||
x3 | x | x | x |
1 | ||
Nie umiesz podstawiś za x+ | =3 | |
x |
1 | 1 | 1 | ||||
x+ | =3 to (x+ | )3= 33=27 i −3*(x+ | )= −3*3=9 | |||
x | x | x |