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13-04-2018
Mathematics

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[tex] {16x}^{2} + 8x + 1[/tex]

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Аноним Аноним

13-04-2018

[tex] {16x}^{2} + 8x + 1 = (4x + 1)(4x + 1) = (4x + 1)^2[/tex]

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[tex]4(x - 1) {}^{2} - 9[/tex]