1)2/[(x-2)(x+2)]-1/[x(x-2)]=(2x-x-2)/[x(x-2)(x+2)]=(x-2)/[x(x-2)(x+2)]=
=1/[x(x+2)]
2)1/[x(x+2)]:1/(x+2)²=1/[x(x+2)]*(x+2)²=(x+2)/x
IABI=√(1+3)²+(-2-y)²)
√(16+(-2-y)²)=5
16+(-2-y)²=25
(-2-y)²=9
1) -2-y=3 2) -2-y=-3
y=-5 y=1
A(-3 ;-5) или А(-3; 1)
<span>10 (n+m)-4(2m+7n) - раскрываем скобки.
10n+10m-8m-28n
2n-18m
</span>