Proszę o rozwiązanie równania:
-4(x²-12) = x²(12-x²)
4(12-x²)-x²(12-x²) = 0
(4-x²)(12-x²) = 0
(2+x)(2-x)(2√3+x)(2√3-x) = 0
2+x = 0 => x = -2
lub
2-x = 0 => x = 2
2√3+x = 0 => x = -2√3
2√3-x = 0 => x = 2√3
Odp. x = -2√3 v x = -2 v x = 2 v x = 2√3
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-4(x²-12) = x²(12-x²)
4(12-x²)-x²(12-x²) = 0
(4-x²)(12-x²) = 0
(2+x)(2-x)(2√3+x)(2√3-x) = 0
2+x = 0 => x = -2
lub
2-x = 0 => x = 2
lub
2√3+x = 0 => x = -2√3
lub
2√3-x = 0 => x = 2√3
Odp. x = -2√3 v x = -2 v x = 2 v x = 2√3