Question

помогите пожалуйста с алгеброй 10 класс ​

Answer 1

ОДЗ:

$\displaystyle\bf\\\left \{ {{x + 2 \geqslant 0} \atop { \frac{2x - 5}{3} \geqslant 0}} \right. \\ \displaystyle\bf\\\left \{ {{x \geqslant - 2} \atop {x \geqslant 2.5 }} \right. \\ \\ x \geqslant 2.5$

$3 \sqrt{x + 2} = 2x - 5 \\ (3 \sqrt{x + 2} ) {}^{2} = (2x - 5) {}^{2} \\ 9(x + 2) = 4 {x}^{2} - 20x + 25 \\ 9x + 18 = 4 {x}^{2} - 20x + 25 \\ 4 {x}^{2} - 20x - 9x + 25 - 18 = 0 \\ 4 {x}^{2} - 29x + 7 = 0 \\ a = 4 \\ b = - 29 \\ c = 7 \\ D = {b}^{2} - 4ac = ( - 29) {}^{2} - 4 \times 4 \times 7 = \\ = 841 - 112 = 729 \: \: ( \sqrt{D} = 27) \\ x_{1} = \frac{29 - 27}{2 \times 4} = \frac{2}{8} = 0.25 \\ x_{2} = \frac{29 + 27}{2 \times 4} = \frac{56}{8} = 7$

Первый корень не подходит

Ответ: х = 7