Question

Решите оба задания, пж!!! Очень срочно надо! Я в математике полный ноль(

Answer 1

$y(0) = 6 \\ y(2) = 16 \\ y( - 1) = 1 \\ \\ {x}^{\frac{2}{3} } = 27 \\ \sqrt[3]{{x}^{2} } =27 \\ {x}^{2} = 19683 \\ x = \binom{ + }{ - } 81 \sqrt{3}$

Answer 2

2) y(x) = 5x + 6

   y(0) = 5 * 0 + 6 = 6

   y(2) = 5 * 2 + 6 = 16

   y( - 1) = 5 * (- 1) + 6 = - 5 + 6 = 1

$3)x^{\frac{2}{3} } =27\\\\\sqrt[3]{x^{2} }=27\\\\(\sqrt[3]{x^{2} })^{3}=27^{3}\\\\x^{2}=27^{2}*3^{2} *3\\\\x_{1}=27*3\sqrt{3}=81\sqrt{3}\\\\x_{2}=-27*3\sqrt{3}=-81\sqrt{3}$