Помогите, пожалуйста

8) $\int\limits^1_0 {(x^3+1)} \, dx = (\frac{x^4}{4} +x)|^1_0= \frac{1}{4} +1-0-0= \frac{5}{4}$

9) Подставляем 4 вместо x
√(4*4 + 9) = √(4 + a) + √4
√25 = √(4 + a) + √4
√(4 + a) = 5 - 2 = 3
4 + a = 9
a = 5

10)
$\left \{ {{3^{3x}=3^{7-y}} \atop { \frac{1}{x}+2 = \frac{12}{y} }} \right.$
$\left \{ {{3x=7-y} \atop { \frac{1+2x}{x} = \frac{12}{y} }} \right.$
$\left \{ {{y=7-3x} \atop { \frac{x}{1+2x} = \frac{y}{12} = \frac{7-3x}{12} }} \right.$
Решаем пропорцию
12x = (1 + 2x)(7 - 3x) = 7 + 11x - 6x^2
6x^2 + x - 7 = 0
(x - 1)(6x + 7) = 0
x1 = 1; y = 7 - 3x = 7 - 3 = 4
x2 = -7/6; y = 7 - 3x = 7 + 7/2 = 21/2 = 10,5