Question

решите уравнение пожалуйста ​

Answer 1

ОДЗ:

$\left \{ {x+5\neq 0} \atop {x-5\neq 0}} \right. =>\left \{ {{x\neq -5} \atop {x\neq 5}} \right.$

$\frac{80}{x+5}+\frac{80}{x-5}=12\\\\\frac{80(x-5)}{(x+5)(x-5)}+\frac{80(x+5)}{(x+5)(x-5)}=12\\\\\frac{80x-400+80x+400}{(x+5)(x-5)}=12\\\\160x= 12(x^{2}-25)\\160x-12x^{2} +12*25=0\\-12x^{2} +160x+300=0\\-6x^{2} +80x+150=0\\-3x^{2} +40x+75=0\\D=b^{2}-4ac=1600-4*(-3)*75=1600+12*75=2500\\\\x1=\frac{-b+\sqrt{D} }{2a}=\frac{-40+{50} }{-6}=-\frac{10}{6}=-\frac{5}{3}=-1\frac{2}{3} \\\\x2=\frac{-b-\sqrt{D} }{2a}=\frac{-40-{50} }{-6}=-\frac{-90}{-6}=15}$

Ответ:

$x=-1\frac{2}{3}\\x=15.$

Answer 2

$\frac{80(x - 5) + 80(x + 5) - 12(x + 5) \times (x - 5)}{(x + 5) \times (x -5) } = 0\\ \frac{80x - 400 + 80x + 400 - 12(x {}^{2} - 25)}{(x + 5) \times (x - 5)} = 0 \\ \frac{160x - 12x {}^{2} + 300 } {(x + 5) \times (x - 5)} = 0 \\ 160x - 12x {}^{2} + 300 = 0 \\ 12 {x}^{2} + 160x + 300 = 0\\ 3x {}^{2} - 5x - 45x - 75 = 0\\ x \times (3x + 5) - 15(3x + 5) = 0\\ (3x + 5) \times (x - 15) = 0 \\ 3x + 5 = 0 \\ x - 15 = 0 \\ x 1= - \frac{5}{3} \\ x2 = 15$