Решите уравнение:
а) $80 + y^2 = 81$;
б) $19 + c^2 = 10$;
в) $20 - b^2 = -5$;
г) $3x^2 = 1,47$;
д) $\frac{1}{4}a^2 = 10$;
е) $-5y^2 = 1,8$.
$80 + y^2 = 81$
$y^2 = 81 - 80$
$y^2 = 1$
$y = ±\sqrt{1}$
y = ±1
$19 + c^2 = 10$
$c^2 = 10 - 19$
$c^2 = -9$ − ∅
$20 - b^2 = -5$
$-b^2 = -5 - 20$
$b^2 = 25$
$b = ±\sqrt{25}$
b = ±5
$3x^2 = 1,47$
$x^2 = 1,47 : 3$
$x^2 = 0,49$
$x = ±\sqrt{0,49}$
x = ±0,7
$\frac{1}{4}a^2 = 10$
$a^2 = 10 : \frac{1}{4}$
$a^2 = 10 * 4$
$a = ±\sqrt{40}$
$a = ±\sqrt{4 * 10}$
$a = ±2\sqrt{10}$
$-5y^2 = 1,8$
$y^2 = 1,8 : (-5)$
$y^2 = -0,36$ − ∅
Пожауйста, оцените решение