| 1 | 1 | |||
Jeśli f(x):R→R oraz dla x≠0,1 zachodzi f(x)+f( | )=(2x−1)2+f(1− | ). Jak polliczyć | ||
| 1−x | x |
| ⎧ | f(3)+f(−1/2)=25+f(2/3) | |
| ⎨ | f(2/3)+f(3)=1/9+f(−1/2) | |
| ⎩ | f(−1/2)+f(2/3)=4+f(3) |
| 1 | 1 | |||
f(x)+f( | )=(2x−1)2+f(1− | ) | ||
| 1−x | x |
| 1 | 2 | |||
f( | )+f(1−1/x)=( | −1)2+f(x) | ||
| 1−x | 1−x |
| 2 | ||
f(1−1/x)+f(x)=(2− | −1)2+f(1/(1−x)) | |
| x |
| x+1 | ||
−a+b+c=( | )2 | |
| x−1 |
| x−2 | ||
a−b+c=( | )2 | |
| x |
| 8 | 8 | |||
Wa=8x2+ | −8x− | +4 | ||
| x2 | x |
| 2 | 2 | |||
f(x)=2x2+ | −2x− | +1 | ||
| x2 | x |