Задание в картинках...

$\frac{1}{ {x}^{2} } + \frac{1}{ {(x + 2)}^{2} } = \frac{10}{9} \\$

ОДЗ:
[ Х =/ 0
[ Х =/ - 2

$\frac{ {x}^{2} + {(x + 2)}^{2} }{ {x}^{2} {(x + 2)}^{2} } = \frac{10}{9} \\ \frac{2 {x}^{2} + 4x + 4}{ {(x(x + 2))}^{2} } = \frac{10}{9} \\ \frac{2( {x}^{2} + 2x) + 4 }{ {( {x}^{2} + 2x) }^{2} } = \frac{10}{9} \\$

Сделаем замену: х^2 + 2х = y , тогда

$\frac{2y + 4}{ {y}^{2} } = \frac{10}{9} \\ 10 {y}^{2} = 18y + 36 \\ 10 {y}^{2} - 18y - 36 = 0 \\ 5 {y}^{2} - 9y - 18 = 0 \\$
D = 81 + 4•5•18 = 81 + 360 = 441

y1 = ( 9 - 21 ) / 10 = - 12 / 10 = - 1,2

y2 = ( 9 + 21 ) / 10 = 30/10 = 3

Обратная замена:

1) х^2 + 2х = - 1,2
х^2 + 2х + 1,2 = 0
D = 4 - 4•1,2 = 4 - 4,8 = - 0,8 , D < 0
Корней нет

2) х^2 + 2х = 3
х^2 + 2х - 3 = 0

Х1 = - 3
Х2 = 1

ОТВЕТ: - 3 ; 1.

x≠0; x≠-2;

$\displaystyle \frac{1}{x^{2} }+\frac{1}{(2+x)^{2}}=\frac{10}{9};\ y=x+1;\ y+1=x+2;\ y-1=x;\\\frac{1}{(y-1)^2}+\frac{1}{(y+1)^{2}}=\frac{10}{9};\\\frac{(y+1)^{2}+(y-1)^{2}}{(y-1)^2(y+1)^{2}}=\frac{10}{9};\\\frac{y^2+y+1+y^2-y+1}{(y^2-1)^2}=\frac{10}{9};\\\frac{y^2+1}{y^4-2y^2+1}=\frac{5}{9};$

9y²+9=5y⁴-10y²+5; t=y²; t>0;

-5t²+19t+4=0;

D=361+4*5*4=441

t=(-19+21)/-10=-0,2; ∅

t=(-19-21)/-10=4;

y=2; y=-2;

y=x+1; x=y-1;

x=1; x=-3;