K²/(k+4)²-16/(-(k+4)²)=k²·(-1)/(k+4)²-16/(-(k+4)²=(-((k²)-16)/(-(k+4)²)=(-1(k-4)(k+4)/(k+4)(k+4)=(-1)(k-4)/(k+4)=(-k+4)/(k+4)
3х-4у=-12
При х=7
3*7-4у=-12
21-4у=-12
-4у=-12-21
-4у=-33
у=-33\-4
у= 8,25
Значит смотри, (sin46-sin14)^2 +(cos46+cos14)^2 = (2sin16*cos30)^2 + (2cos30*cos16)^2 = (2sin16*√3/2)^2 + (2*√3/2*cos16)^2 =(двойки сократились) sin^2 16*3 + 3*cos^2 16 = 3(sin^2 16 + cos^2 16) = 3