т.к. y стоит в знаменателе ставим ограничение что y не равен 0
![g( x^{2} -4) = \frac{1}{ x^{2} -4}\\\\f(x+2)= \frac{2}{x+2}\\\\ \frac{1}{ x^{2} -4}- \frac{2}{x+2} =-1\\\\ \frac{1}{(x-2)(x+2)}- \frac{2}{x+2}+1=0\\\\ \frac{1-2x+4+ x^{2} -4}{ x^{2} -4} =0\\\\ \frac{ x^{2} -2x+1}{ x^{2} -4}=0](https://tex.z-dn.net/?f=g%28+x%5E%7B2%7D+-4%29+%3D++%5Cfrac%7B1%7D%7B+x%5E%7B2%7D+-4%7D%5C%5C%5C%5Cf%28x%2B2%29%3D+%5Cfrac%7B2%7D%7Bx%2B2%7D%5C%5C%5C%5C+%5Cfrac%7B1%7D%7B+x%5E%7B2%7D+-4%7D-+%5Cfrac%7B2%7D%7Bx%2B2%7D+%3D-1%5C%5C%5C%5C+%5Cfrac%7B1%7D%7B%28x-2%29%28x%2B2%29%7D-+%5Cfrac%7B2%7D%7Bx%2B2%7D%2B1%3D0%5C%5C%5C%5C+%5Cfrac%7B1-2x%2B4%2B+x%5E%7B2%7D+-4%7D%7B+x%5E%7B2%7D+-4%7D+%3D0%5C%5C%5C%5C+%5Cfrac%7B+x%5E%7B2%7D+-2x%2B1%7D%7B+x%5E%7B2%7D+-4%7D%3D0++++++)
x² - 2x + 1 = 0 x ≠ 2 x ≠ - 2
(x - 1)² = 0
x = 1
Домножим первое уравнение на -х: х^2-ху=-3х
Сложим уравнения:
х^2-ху+-ху=10-3х
х^2+3х-10=0
D=9+40=49
x=(-3+7)/2.
x=(-3-7)/2
x=2;-5
y=-2;5 Ответ:(2;5);(-5;-2)
4( 3m - n ) - 3( 4m + 2n ) = 12m - 4n - 12m - 6n = - 10n
Б)x+4/7=x-9/8
8(x+4)=(x-9)7
8x+32=7x-63
8x-7x=-63-32
<span>x=-95
а)2x-3=4x-11
4x-2x=-3+11
2x=8
x=8/2
x=4</span>