| 222 + 221 | 211(211 + 210) | 211 + 210 | ||||
x = | = | = | ||||
| 3*211 | 3*211 | 3 |
| 28 − 43 | 28 − (22)3 | 28−26 | ||||
x = | = | = | = | |||
| 162+82 | (24)2 + (23)2 | 28+26 |
| 26(22−1) | 3 | |||
= | ||||
| 26(22+1) | 5 |
| n+2 | ||
n=1∑oo | ||
| 2n3−1 |
| 1 | ||
Odpowiedź: zbieżny. Wskazówka: porównać z szeregiem n=1∑00 | ||
| n2 |
| 1 | ||
n=1∑oo | ||
| 2n − 1 |
| 4x(x+3) | |
| 8x(x−2) |
| x+3 | ||
czy ja te 4 i 8 mogę normalnie skrócić, tak że końcowy wynik będzie równał się | ||
| 2x(x−2) |
| ||||||||
F(ax3+bx2+cx+d)=a*A3+b*A2+c*A+d*I2x2, gdzie A= | . | |||||||
| √3−1 | 9−5√3 | |||
Wykaz ze : | =p3{ | |||
| √3+1 | 9+5√3 |
| 1 | ||
Wykaż, że jeśli a < −2 i b > | to 2ab < a + 2 − 4b | |
| 2 |
| 2 | ||
Ktora z liczb a=(16*0,253n−2):(−0.5*0,253n−4) czy b=81*( | )12−4n | |
| 3 |
| 2 | ||
*3*(− | )4n−7 gdzie n nalezy do liczb nnaturalnych i o ile ? | |
| 3 |
| a3/2+b3/2 | a−2/3 3√a−b | |||
Obliczyc wartosc wyrazenia | : | |||
| (a2−ab)2/3 | a√a−b√b |
| 3 | ||
dla a=1,2 b= | ||
| 5 |
| a3/2+b3/2 | |
| (a(a−b))2/3 |
| a3/2−b3/2 | a3−b3 | |||
* | = | = | ||
| a−2/3*3√a−b | a2/3*3√(a−b)2*a−2/3*√a−b |
| a3−b3 | a3−b3 | |||
= | = | = a2+ab+b2 | ||
| 3√(a−b)3 | a−b |
| n+2+√n2−4 | n+2−√n2−4 | ||
+ | |||
| n+2−√n2−4 | n+2+√n2−4 |
| n3 | ||
limn−>∞ | ||
| 2n |
| (2014+(−1)n)n | ||
∑ | *(x+1)n Obliczyłem promień zbieżności, R=1/2015. Jednak | |
| n |
| x0,5+1 | 1 | ||
: | |||
| x+x1/2+1 | x1,5−1 |
| x1/2+1 | |
* x3/2−1= | |
| x+x1/2+1 |
| x1/2+1 | |
* (x1/2)3−13 | |
| x+x1/2+1 |
| x1/2+1 | |
*(x1/2−1)(x+x1/2+1)=(x1/2+1)(x1/2−1)= x−1 | |
| x+x1/2+1 |
| p | r | p | ||||
Jeśli p, q, r, s są liczbami dodatnimi takimi, że | < | , to | < | |||
| q | s | q |
| p + r | p + r | r | ||||
oraz | < | |||||
| q + s | q + s | s |
| b | Δ | |||
y= a[(x+ | )2 − | ] | ||
| 2a | 4a2 |
| x2 | ||
2x+√x2−1(1+ | ) | |
| x2−1 |
| √x2−1*x2 | ||
2x+√x2−1+ | = po sprowadzeniu do wspolnego mianownika | |
| x2−1 |
| 2x(x2−1) +√x2−1(x2−1)+√x2−1*x2 | |
= | |
| x2−1 |
| 2x3−2x+√x2−1x2−√x2−1+√x2−1x2 | |
= | |
| x2−1 |
| 2x3−2x+2√x2−1*x2−√x2−1 | |
mozna cos jeszce z tego wyczarowac ? | |
| x2−1 |
| 1 | ||
A potem jeszce mam od tego odjac | ||
| x2−1 |