<span>y^2 - (11y - 2) /9 =0</span>
9y^2 - 11y + 2 = 0
Δ = 121 - 72 = 49 = 7^2
y1 = (11+7)/18 = 18/18 = 1
y2 = (11-7)/18 = 4/18 = 2/9
X² - 6x + 29 = x² - 2*3x + 3² - 3² + 29 = (x-3)² -9 + 29 = (x-3)² + 20;
(x-3)² ≥ 0
x - 3 ≥ 0
x ≥ 0
(x-3)² + 20 ≥ 0
Надеюсь, помог.
ОТВЕТ:
y(x) = (2/3)*x^3 + (1/2)*x^2 - 1/2
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РЕШЕНИЕ:
для функции
f(x) = 2x^2 + x
первообразная есть:
y(x) = (2/3)*x^3 + (1/2)*x^2 + С,
подставив А (1;1), найдем
1 = (2/3)*1^3 + (1/2)*1^2 + C, откуда
C = 1 - 2/3 - 1/2 =(6 -4 - 3)/6= - 1/2
<span>y(x) = (2/3)*x^3 + (1/2)*x^2 - 1/2</span>