AM+CM | ||
BM= | ||
2 |
AM+x | ||
a−x= | ||
2 |
2 | ||
2a−3x≥0 ⇒x< | a | |
3 |
12a−√a2+2b2 | ||
x1= | ||
16 |
a−b | a | b | ||||
nawet z ebym skorzystal ze wzoru | = | − | to dostane x1= | |||
c | c | c |
12a | ||
−√a2+2b2}{16}= | ||
16 |
3a | √a2+2b2 | |||
= | − | |||
4 | 16 |
3a−√a2+b2 | ||
x1= | ||
4 |
12a−√a2+2b2 | ||
To tak x1= | ||
16 |
12a+√a2+2b2 | ||
a x2= | ||
16 |
2 | ||
Ale dla x1 i x2 jest x∊(0, | a) Teraz muszse sprawdzic warunki rozwaizalnosci | |
3 |