Предмет: Математика
Автор: barvinkalev
Вопрос задан 1 год назад

Помогите пожалуйста решить уравнения,!!!

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Ответы

Ответ дал: sebrithien

$\frac{3}{x - 19} - \frac{19}{x - 3} = 0 \\ \frac{3(x - 3) - 19(x - 19)}{(x - 19)(x - 3)} = 0 \\ \frac{3x - 9 - 19x + 361}{(x - 19)(x - 3)} \\ x - 19 ≠0 \\ x≠19 \\ x - 3≠0 \\ x≠3 \\ - 16x + 352 = 0 \\ - 16x = - 352 \\x = - \frac{352}{16} \\ x = 22$

$\frac{5x + 4}{2} + 3 - \frac{9x}{4} = 0 \\ \frac{6(5x + 4) + 36 - 27x}{12} = 0 \\ \frac{30x + 24 + 36 - 27x}{12} = 0 \\ 3x + 60 = 0 \\ 3x = - 60 \\ x = - \frac{60}{3} \\ x = - 20$

$3 - \frac{x}{7} - \frac{x}{3} = 0 \\ \frac{63 - 3x - 7x}{21} = 0 \\ - 10x = 63 \\ x = - \frac{63}{10} \\ x = - 6.3$

$\frac{x - 6}{2} - \frac{x}{3} - 3 = 0 \\ \frac{3(x - 6) - 2x - 18}{6} = 0 \\ 3x - 18 - 2x - 18 = 0 \\ x - 36 = 0 \\ x = 36$

$\frac{9}{x - 2} - \frac{9}{2} = 0 \\ \frac{18 - 9(x - 2)}{2(x - 2)} = 0 \\ x - 2≠0 \\ x ≠2 \\ 18 - 9x + 18 = 0 \\ - 9x + 36 = 0 \\ - 9x = - 36 \\ x = \frac{ - 36}{ - 9} \\ x = 4$

$x - \frac{6}{x} + 1 = 0 \\ \frac{ {x}^{2} - 6x + x}{x} = 0 \\ x≠0 \\ {x}^{2} - 6x + x = 0 \\ D = {( - 6)}^{2} - 4 \times 1 \times 1 = 36 - 4 = 32 \\ \\ \sqrt{D} = \sqrt{32} = \sqrt{16 \times 2} = 4 \sqrt{2} \\ x1 = \frac{6 + 4 \sqrt{2} }{2} = \frac{2(3 + 4 \sqrt{2}) }{2} = 3 + 4 \sqrt{2} \\ x2 = \frac{6 - 4 \sqrt{2} }{2} = \frac{2(3 - 4 \sqrt{2} )}{2} = 3 - 4 \sqrt{2}$