Question

(4x-3)×(x+1)=0; 16-x^2:x+4=0 решите умоляяяяююююю даю 20 баллов

Answer 1

Ответ:

1) (4x - 3)(x + 1) = 0

4х² + 4х - 3х - 3 = 0

4х² + х - 3 = 0

D = 1 + 48 = 49

$x_1 = \frac{-1-7}{8} = -\frac{8}{8} = -1\\  \\x_2 = \frac{-1+7}{8} = \frac{6}{8} = \frac{3}{4} = 0,75$  

$2) \frac{16-x^2}{x+4} = 0\\\\\frac{(4-x)(4+x)}{x+4} = 0\\ \\4 - x = 0\\\\x = 4$

Answer 2

$(4x-3)(x+1)=0\\\\\left[\begin{array}{ccc}4x-3=0\\x+1=0\end{array}\right\\\\\left[\begin{array}{ccc}4x=3\\x=-1\end{array}\right\\\\\left[\begin{array}{ccc}x=0,75\\x=-1\end{array}\right\\\\Otvet:\boxed{0,75;-1}$

$2)\frac{16-x^{2}}{x+4}=0\\\\\left \{ {{16-x^{2}=0 } \atop {x+4\neq0 }} \right.\\\\\left \{ {{(4-x)(4+x)=0} \atop {x\neq-4 }} \right. \\\\4-x=0\\\\x_{1}=4\\\\4+x=0\\\\ x_{2}=-4\\\\Otvet:\boxed{4}$