Помогите пожалуйста 4,5.Буду очень благодарна

4)
$a + b = - 2 {x}^{3} y - 1.5x {y}^{2} - 0.4 {x}^{3} y + 1.5x {y}^{2} = - 2.4 {x}^{3} y$
при х = 1,5; y = -2:
$- 2.4 \times {1.5}^{3} \times ( - 2) = - \frac{24}{10} \times { (\frac{3}{2}) }^{3} \times ( - 2) = \frac{24}{5} \times \frac{27}{8} = \frac{81}{5} = 16.2$
$a - b = - 2 {x}^{3} y - 1.5x {y}^{2} - ( - 0.4 {x}^{3} y + 1.5x {y}^{2} ) = - 2 {x}^{3} y - 1.5 x{y}^{2} + 0.4 {x}^{3} y - 1.5x {y}^{2} = - 1.6 {x}^{3} y - 3x {y}^{2}$
при х = 1,5; у = -2:
$- 1.6 \times ( \frac{3}{2} ) ^{3} \times ( - 2) - 3 \times \frac{3}{2} \times {( - 2)}^{2} = - \frac{16}{10} \times \frac{27}{8} \times ( - 2) - \frac{9}{2} \times 4 = \frac{2 \times 27}{5} - 9 \times 2 = \frac{54}{5} - 18 = 10.8 - 18 = - 7.2$
5)
$2( x + 3)(x - 2) - (2x + 1)(x - 3) - 7 = 0 \\ 2( {x}^{2} + 3x - 2x - 6) - (2 {x}^{2} + x - 6x - 3) = 0 \\ 2 {x}^{2} + 2x - 12 - 2 {x}^{2} + 5x + 3 = 0 \\ 7x - 9 = 0 \\ 7x = 9 \\ x = \frac{9}{7} = 1 \frac{2}{7}$