Question

помогите два лёгких задание на это можно потратить 5 мин​

Answer 1

Ответ:

1.

$3x(2x + 1) - x(6x - 1) = 10 \\ 6 {x}^{2} + 3x - 6 {x}^{2} + 1 = 10 \\ 3x = 10 - 1 \\ 3x = 9 \\ x = 9 \div 3 \\ x = 3$

$\frac{x - 1}{2} - \frac{x + 1}{3} = 1 \\ \frac{(x - 1) \times 3 - (x + 1) \times 2}{6} = 1 \\ \frac{3x - 3 - 2x - 2}{6} = 1 \\ x - 5 = 6 \\ x = 6 + 5 \\ x = 11$

$4 {x}^{2} - 3x = 0 \\ x(4x - 3) = 0 \\ 1)x = 0 \\ 2)4x - 3 = 0 \\ 4x = 3 \\ x = \frac{3}{4}$

$\frac{6x - 1}{5} - \frac{2 - x}{4} = \frac{3x + 2}{2} \\ \frac{(6x - 1) \times 4 - (2 - x) \times 5}{20} = \frac{(3x + 2) \times 10}{20} \\ 24x - 4 - 10 + 5x = 30x + 20 \\ 29x - 30x = 20 + 14 \\ - x = 34 \\ x = - 34$

$x(4x - 2) - 2x(2x + 4) = 4 \\ 4 {x}^{2} - 2x - 4 {x}^{2} - 8x = 4 \\ - 10x = 4 \\ x = - 4 \div 10 \\ x = - 0.4$

$\frac{3x}{5} - \frac{x + 1}{2} = 1 \\ \frac{3x \times 2 - (x + 1) \times 5}{10} = 1 \\ \frac{6x - 5x - 5}{10} = 1 \\ x - 5 = 10 \\ x = 10 + 5 \\ x = 15$

$2 {x}^{3} - 5x = 0 \\ x(2 {x}^{2} - 5) = 0 \\ 1)x = 0 \\ 2)2 {x}^{2} - 5 = 0 \\ 2 {x}^{2} = 5 \\ {x}^{2} = 5 \div 2 \\ {x}^{2} = 2.5 \\ {x}^{2} -2.5 = 0 \\(x - \sqrt{2.5})(x + \sqrt{2.5} ) = 0 \\ 2.1) x - \sqrt{2.5} = 0 \\ x = \sqrt{2.5} \\ 2.2)x + \sqrt{2.5} = 0 \\ x = - \sqrt{2.5}$

$\frac{2x - 3}{6} - \frac{4 - x}{3} = \frac{x - 1}{2} \\ \frac{2x - 3 - (4 - x) \times 2}{6} = \frac{(x - 1) \times 3}{6} \\ 2x - 3 - 8 + 2x = 3x - 3 \\ 4x - 3x = - 3 + 11 \\ x = 8$