Question

Помогите решить второй
Дам много балов​

Answer 1

Ответ:

2) при $x\in \{0;\:2 \frac{3}{19} \}$

Объяснение:

$\frac{3 {x}^{2} + 7x }{4} = \frac{7 {x}^{2} - 5x }{3} \: \: \: \: \bigg | \small \times 12 \\ 3(3 {x}^{2} + 7x) = 4(7 {x}^{2} - 5x) \\ 9 {x}^{2} + 21x = 28 {x}^{2} - 20x \\ 28 {x}^{2} - 9 {x}^{2} - 20x - 21x = 0 \\ 19 {x}^{2} - 41x = 0 \\ x(19x - 41) = 0 \: \\ \left[ \begin{array}{l}x = 0 \\ 19x - 41 = 0 \end{array} \right. \: < = > \left[ \begin{array}{l}x = 0 \\ x = \dfrac{41}{19} =2 \frac{3}{19} \end{array} \right.$