(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
(a + b)(a-b) = a2 – b2
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(a – b)3 = a3 – 3a2b + 3ab2 – b3
a3 – b3 = (a – b)(a2 + ab + b2)
a3 + b3 = (a + b)(a2 – ab + b2)
a1/n = $\sqrt[n]{a}$ np. 81/3 = $\sqrt[3]{8}$ = 2
am/n = (a1/n)m = ($\sqrt[n]{a}$)m lub $\sqrt[n]{a^{m}}$
np. 163/2 = (161/2)3 = ($\sqrt{16}$)3 = 43 = 64
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
(a + b)(a-b) = a2 – b2
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(a – b)3 = a3 – 3a2b + 3ab2 – b3
a3 – b3 = (a – b)(a2 + ab + b2)
a3 + b3 = (a + b)(a2 – ab + b2)
a1/n = $\sqrt[n]{a}$ np. 81/3 = $\sqrt[3]{8}$ = 2
am/n = (a1/n)m = ($\sqrt[n]{a}$)m lub $\sqrt[n]{a^{m}}$
np. 163/2 = (161/2)3 = ($\sqrt{16}$)3 = 43 = 64
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
(a + b)(a-b) = a2 – b2
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(a – b)3 = a3 – 3a2b + 3ab2 – b3
a3 – b3 = (a – b)(a2 + ab + b2)
a3 + b3 = (a + b)(a2 – ab + b2)
a1/n = $\sqrt[n]{a}$ np. 81/3 = $\sqrt[3]{8}$ = 2
am/n = (a1/n)m = ($\sqrt[n]{a}$)m lub $\sqrt[n]{a^{m}}$
np. 163/2 = (161/2)3 = ($\sqrt{16}$)3 = 43 = 64
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
(a + b)(a-b) = a2 – b2
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(a – b)3 = a3 – 3a2b + 3ab2 – b3
a3 – b3 = (a – b)(a2 + ab + b2)
a3 + b3 = (a + b)(a2 – ab + b2)
a1/n = $\sqrt[n]{a}$ np. 81/3 = $\sqrt[3]{8}$ = 2
am/n = (a1/n)m = ($\sqrt[n]{a}$)m lub $\sqrt[n]{a^{m}}$
np. 163/2 = (161/2)3 = ($\sqrt{16}$)3 = 43 = 64
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
(a + b)(a-b) = a2 – b2
(a + b)3 = a3 + 3a2b + 3ab2 + b3
(a – b)3 = a3 – 3a2b + 3ab2 – b3
a3 – b3 = (a – b)(a2 + ab + b2)
a3 + b3 = (a + b)(a2 – ab + b2)
a1/n = $\sqrt[n]{a}$ np. 81/3 = $\sqrt[3]{8}$ = 2
am/n = (a1/n)m = ($\sqrt[n]{a}$)m lub $\sqrt[n]{a^{m}}$
np. 163/2 = (161/2)3 = ($\sqrt{16}$)3 = 43 = 64
o ile % więcej = różnica pomiędzy ilościami przez większą ilość razy 100%
x-liczba , a-przybliżenie , błąd bezwzględny |x-a|, błąd względny $\frac{|x - a|}{|x|}$