Question

Помогите с 3, 4 и 5 заданиями, пожалуйста​

Answer 1

Ответ:

3.

$1 \\ ∫ (4x ^{3} - 2) ^{4} x ^{2} dx\\ 0$

$∫(4 {x}^{3} - 2) ^{4} x ^{2} dx$

$∫(4 {x}^{3} - 2) ^{4} {x}^{2} \times \frac{1}{4 \times 3 {x}^{2} } dt$

$∫ {(4 {x}^{3} - 2)}^{4} \times \frac{1}{4 \times 3} dt$

$∫ {(4 {x}^{3} - 2)}^{4} \times \frac{1}{12} dt$

$∫ \frac{(4 {x}^{3} - 2) ^{4} }{12} dt$

$∫ \frac{ {t}^{4} }{12} dt$

$\frac{1}{12} \times ∫ {t}^{4} dt$

$\frac{1}{12} \times \frac{ {t}^{4 + 1} }{4 + 1}$

$\frac{1}{12} \times \frac{ {t}^{5} }{5}$

$\frac{1}{12} \times \frac{(4 {x}^{3} - 2) ^{5} }{5}$

$\frac{(4 {x}^{3} - 2) ^{5} }{60}$

$\frac{(4 {x}^{3} - 2) ^{5} }{60} 1 |0$

$\frac{(4 \times {1}^{3} - 2) ^{5} }{60} - \frac{(4 \times {0}^{3} - 2) ^{5} }{60}$

$\frac{32}{60} - \frac{ - 32}{60}$

$\frac{16}{15}$

4.

$x + y - z = 1 \\ 2x - y + z = 5 \\ x + y - 2z = - 1$

$x + y - z = 1 \\ 2x - y + z = 5 \\ \\ 2x - y + z = 5 \\ x + y - 2z = - 1$

$x + y - z1 \\ 2x - y + z = 5$

$3x = 6$

|:3

$x = 2$

$2x - y + z = 5 \\ x + y - 2z = - 1$

$2x - y + z + x + y - 2z = 5 - 1$

$3x - z = 4$

$x = 2 \\ 3x - z = 4$

$3 \times 2 - z = 4$

$z = 2$

$z = 2 \\ x = 2$

$2 + y - 2 = 1$

$y = 1$

$(x.y.z) = (2.1.2)$

$2 + 1 - 2 = 1 \\ 2 \times 2 - 1 + 2 = 5 \\ 2 + 1 - 2 \times 2 = - 1$

$1 = 1 \\ 5 = 5 \\ - 1 = - 1$

$(x.y.z) = (2.1.2)$

5....