Question

4 задание, фото ниже

Answer 1

$\frac{10}{25-x^2} - \frac{1}{5+x} - \frac{x}{x-5} =0\; ,\; \; \; \; OOF:\; x\ne \pm 5\\\\ \frac{10}{(5-x)(5+x)} - \frac{1}{5+x} + \frac{x}{5-x} =0\\\\ \frac{10-5+x+5x+x^2}{(5-x)(5+x)} =0\\\\ \frac{x^2+6x+5}{(5-x)(5+x)} =0\; ,\; \; \frac{(x+1)(x+5)}{(5-x)(5+x)} =0\; ,\; \frac{x+1}{5-x} =0\\\\x=-1$

Answer 2

25-x²=(5-x)(5+x),
(x-5)=(-(5-x)
(10-(5-x)+x(5-x))/(5-x)(5+x)= 0
(10-5-x+5x-x²)/(5-x)(5+x)=0
5+4x-x²=0 *(-1)
x²-4x-5=0,(a=1,b=-4,c=-5)
D=b²-4ac, D=16+20=36=6²
x1,x2=(-b+-корень изD)/2a
x1=(4+6)/2=5
x2=(4-6)/2= -1