(3x+4y)в квадрате - (4y-3x)в квадрате =
=((4y+3х)-(4y-3x))((4y+3х)+(4y-3x))=
=(4y+3х-4y+3x)(4y+3х+4y-3x)=6x*8y=48ху
Ответ:
Формула суммы арифметической прогрессии.
3^(x^2 + y^2) = 81
log 2(x) + 2log 4(y) = 1
<span>3^(x^2 + y^2) = 3^3 </span>
log 2(x) + 2log (2^2) (y) =1
<span>x^2 + y^2 = 3 </span>
log 2(x) + 2/2 * log 2(y) = 1
<span>x^2 + y^2 = 3 </span>
log 2(xy) = 1
<span>x^2 + y^2 = 3 </span>
2^1 = xy
<span>x^2 + y^2 = 3 </span>
xy = 2
<span>x = 2/y </span>
(2/y) * y = 2
<span>2 = 2 => x и у - любые числа, не = 0</span>
3a=6i+9j-3k
-2b=-8i+2j-4k
c=6i-8i+9j+2j-3k-4k=-2i+11j-7k