Question

ХЕЕЕЕЛП 1-6 ЗАДАНИЯ

Answer 1

$b) \frac{5}{x} + \frac{4}{x - 3} = 3 \\ 5x - 15 + 4x = 3 \\ 9x = 18 \\ x = 2 \\ 4) {x}^{2} - 7x < 0 \\ x(x - 7) < 0 \\ x < 0 \\ x < 7 \\ ( {x }^{2} - 16)(x + 4)(x + 16) \geqslant 0 \\ {x}^{4} + 4 {x}^{3} - 16 {x}^{2} - 64x + 16 {x}^{3} + 64 {x}^{2} - 256x - 1024$
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