2^6=64
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<h3><em>Решение приложено</em></h3>
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ОДЗ:
По т. Виета
Приравняем к нулю
<em>______+______(1)_____-______(10)_____+______>
</em>x ∈ (1;10)
Наименьшее целое число это 2.
Ответ:
(5x - 1)^2 - 16x^2 = 0
(5x - 1 - 4x)(5x - 1 + 4x) = 0
(x - 1)(9x - 1) = 0
x - 1 = 0
x = 1
9x - 1 = 0
9x = 1
x = 1/9
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x^2 + 6x + 9 + x^2 - 8x + 16 = 2(4x + 12 - x^2 - 3x)
2x^2 - 2x + 25 = 2(x + 12 - x^2)
2x^2 - 2x + 25 = 2x + 24 - 2x^2
4x^2 - 4x + 1 = 0
(2x - 1)^2 = 0
2x - 1 = 0
2x = 1
x = 1/2 = 0,5
Ответ:
a²-49 = (a - 7)(a + 7)
x²-0,81 = (x - 0,9)(x + 0,9)
256-a⁴ = (16 - a²)(16 + a²)
a³+125 = (a + 5)(a² - 5a + 25)
64 - b³ = (4 - b)(16 + 4b + b²)