Question

Решыте пожалуйста , заранее спасибо !

Answer 1

1)

$y = \frac{(x + 3)(x + 7)}{x + 5} \\ x + 5\neq0 \\ x\neq - 5 \\ x \: \epsilon\: ( -\propto; \: - 5 )U( - 5; \: + \propto) \\ \\ b) \: y = 0 \\ \frac{(x + 3)(x + 7)}{x + 5} = 0 \\ (x + 3)(x +7 ) = 0 \\ x_{1} = - 3 \\ x_{2} = - 7 \\ \\ v) \: x = 0 \\ y = \frac{(0 + 3)(0 + 7)}{0 + 5} = \frac{3 \times 7}{5} = \frac{21}{5} = 4.2 \\otvet \: \: \: (0; \: 4.2)$

2)

$f(x) = \frac{x}{6} + 1 \\ f(12) = \frac{12}{6} + 1 = 2 + 1 = 3 \\ f(0) = \frac{0}{6} + 1 = 0 + 1 = 1 \\ f( - 2) = \frac{ - 2}{6} + 1 = - \frac{1}{3} + 1 = \frac{ - 1 + 3}{3} = \frac{2}{3}$

3)

$y = {x}^{4} \\ \\1) \: y = 16 \\ {x}^{4} = 16 \\ ( {x}^{2} ) = {4}^{2} \\ \\ x {}^{2} = 4 \\ {x}^{2} = - 4 \: \: \: \: net \: \: \: korney \\ \\ {x}^{2} = {2}^{2} \\ x_{1} = - 2\\ x_{2} = 2 \\ \\ 2) \: y = \frac{1}{81} \\ {x}^{4} = \frac{1}{81} \\ ( {x}^{2}) {}^{2} = (\frac{1}{9} ) {}^{2} \\ \\ x {}^{2} = - \frac{1}{9} \: \: \: \: net \: \: \: korney \\ x {}^{2} = \frac{1}{9} \\ {x}^{2} = ( \frac{1}{3} ) {}^{2} \\ x_{1} = - \frac{1}{3} \\ x_{2} = \frac{1}{3} \\ \\ 3) \: y = 0 \\ {x}^{4} = 0 \\ x = 0 \\ \\ 4) \: y = - 625 \\ {x}^{4} = - 625 \\ net \: \: \: korney$