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sega963
@sega963
June 2021
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On considère la fonction f : x fonction de x² +1
1) Quels sont les antécédents par f de 5? de 17 ? de 82 ?
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nonotata
X²+1 = 5
x²+1-5 =0
x²-4 =0
x²-2² =0
(x-2)(x+2) =0
donc
x-2 =0 x+2 =0
x = 2 x = -2
x²+1 =17
x²+1-17 =0
x²-16 =0
x²-4² =0
(x-4)(x+4) =0
donc
x-4 = 0 x+4 =0
x = 4 x = -4
x² +1 = 82
x²+1-82 =0
x²-81 =0
x²-9² =0
(x-9)(x+9) =0
donc
x-9 =0 x+9 =0
x = 9 x = -9
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x²+1-5 =0
x²-4 =0
x²-2² =0
(x-2)(x+2) =0
donc
x-2 =0 x+2 =0
x = 2 x = -2
x²+1 =17
x²+1-17 =0
x²-16 =0
x²-4² =0
(x-4)(x+4) =0
donc
x-4 = 0 x+4 =0
x = 4 x = -4
x² +1 = 82
x²+1-82 =0
x²-81 =0
x²-9² =0
(x-9)(x+9) =0
donc
x-9 =0 x+9 =0
x = 9 x = -9