Janek191:
1 | | 1 | | 1 | |
| + |
| = |
| , x ≠ 0 i y≠ 0 |
x | | y | | 12 | |
x + y = 50 ⇒ y = 50 − x
x*(50 − x) = 600
− x
2 + 50 x − 600 = 0
x
2 − 50 x + 600 = 0
Δ = 2 500 = 4*1*600 = 100
√Δ = 10
| 50 − 10 | | 50 + 10 | |
x = |
| = 20 lub x = |
| = 30 |
| 2 | | 2 | |
więc
y = 50 − 20 = 30 lub y = 50 − 30 = 20
Odp. ( 20, 30) , ( 30, 20)
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